14440350 Buffer Hydrolysis and Solubility Solubility Product Worksheet by Hariyanto SMA Negeri 3 Malang

(worksheet 1st)

Drs. M. Hariyanto SMA Negeri 3 Malang

(Public Senior High School) Jl. Sultan Agung Utara No 7 Phone (0341) 324 768 Malang 65111

(web site : http://www.har-chemist.co.nr )

Before we study about buffer solution, we should review the previous material, how to calculate the pH of acid, base, the mixture of acid and base, and Bronsted-Lowry theory of acid-base. Complete the following table!

1. pH of acid and base

Solutions SA/WA/SB/WB The formulae of [H+] or [OH-]

Calculation of pH

H2SO4 0,005 M

SA [H+] = a. Ma …………..…………..

KOH 0,1 M

………….. ………….. …………..…………..

HCOOH 0,1 M Ka =1 x 10-5

………….. ………….. …………..…………..

NH3 0,1 M Kb = 1 x 10-5

………….. ………….. …………..…………..

SA = strong acid; WA = weak acid; SB = strong base; WB = weak base

2. pH of mixture

No The mixtures React or not The steps of solutions 1 10 mL of H2SO4 0.1 M is

added to 90 mL of water. Calculate the pH of mixture!

Not react Dilution ……………………………………… ……………………………………… ………………………………………

2 800 mL of H2SO4 0.1 M is added to 200 mL of HCl 0.2M. Calculate the pH of mixture!

……………………………………… ……………………………………… ………………………………………

3 800 mL of Ca(OH)2 0.1 M is added to 200 mL of NaOH 0.2M. Calculate the pH of mixture!

……………………………………… ……………………………………… ………………………………………

4 800 mL of Ca(OH)2 0.1 M is added to 200 mL of HNO3 0.2M. Calculate the pH of mixture

……………………………………… ……………………………………… ……………………………………… ……………………………………… ……………………………………… ………………………………………

Worksheet of chemistry/xi krsnbi/ by m. hariyanto page 1

http://www.har-chemist.co.nr/

3. The acid dissociation constant (Ka) and base dissociation constant (Kb) a. The ionization of weak acid, HA

1) write the ionization of weak acid, HA in water 2) write the formulae of Ka

b. The ionization of weak base, LOH 1) write the ionization of weak base, LOH in water 2) write the formulae of Kb

4. The Bronsted-Lowry Theory of acid-base

a. Determine the conjugate-base of acid and conjugate-acid of base Complete the blank column!

No Acid Conjugate-base + H+ 1 H2O 2 CH3COOH 3 HCN 4 NH4

+

No Base + H+ Conjugate-acid 1 H2O + H+ 2 NH3 3 HCO3

- 4 HPO4

2- b. Given that the reaction : CO3

2-(aq) + NH4+(aq) ↔ HCO3

-(aq) + NH3(aq) Mention the Bronsted-Lowry acid and base, conjugate pairs of the equation above:


How to make buffer solutions?

A buffer solution is one which resists changes in pH when small quantities of an acid or an alkali are added to it. Based on the value of pH, buffer is classified into two kinds of buffer, an acidic buffer solution is simply one which has a pH less than 7 (pH of 4.76), and An alkaline buffer solution has a pH greater than 7 ( pH of 9.25)

a. Acidic buffer solution

Please observe the following data! No Solutions / mixtures The kind of each solution Acidic buffer /

alkaline buffer / not buffer

1 CH3COOH (aq) + CH3COO- (aq) CH3COOH (aq) = weak acid CH3COO- (aq) = conjugate-base

Acidic buffer

2 HCN (aq) + CN- (aq) …………………………….. ……………………………..

Acidic buffer

3 CH3COOH (aq) + CH3COONa (aq) CH3COOH (aq) = weak acid CH3COO- (aq) = the salt of CH3COOH

Acidic buffer

4 HCl (aq) + Cl- (aq) …………………………….. ……………………………..

Not buffer

5 HCl (aq) + NaCl (aq) …………………………….. ……………………………..

Not buffer

b. Alkaline buffer solution

No Solutions / mixtures The kind of each solution Acidic buffer / alkaline buffer / not buffer

1 NH3 (aq) + NH4+(aq) ……………………………..

……………………………..

Alkaline buffer

2 NH3 (aq) + NH4Cl (aq) …………………………….. ……………………………..

Alkaline buffer

3 NaOH(aq) + NaCl (aq) …………………………….. ……………………………..

Not buffer

CONCLUSION Based on the data above, conclude how to make the acidic buffer solutions and alkaline buffer

solutions?


EXERCISE 1 01. Identify the following mixtures, and classify into acidic buffer solution / alkaline buffer

solution / not buffer! a. HCOOH (aq) + HCOO-(aq) b. HCOOH (aq) + (HCOO)2Ca(aq) c. H2SO4(aq) + HSO4

-(aq) d. H3PO4(aq) + H2PO4

-(aq) e. KOH(aq ) + KCl(aq f. NH3(aq) + (NH4)2SO4(aq)

02. Determine the kind of substance that remains, and classify into acidic buffer solution / alkaline buffer solution / not buffer!

a. 100 mL of ethanoic acid solution 0.1M is added into 50 mL of sodium hydrokxide solution 0.1 M.

b. 100 mL of ethanoic acid solution 0.1M is added into 50 mL of calcium hydrokxide solution 0.1 M.

c. 100 mL of ammonia solution 0.1M is added into 50 mL of sulfuric acid solution 0.1 M.

d. 100 mL of ammonia solution 0.1M is added into 50 mL of nitric acid solution 0.1 M.


This hand out describes simple acidic and alkaline buffer solutions and explains how they work.

What is a buffer solution?

Definition

A buffer solution is one which resists changes in pH when small quantities of an acid or an alkali are added to it.

Acidic buffer solutions

An acidic buffer solution is simply one which has a pH less than 7. Acidic buffer solutions are commonly made from a weak acid and its conjugate-base or and one of its salts - often a sodium salt,

A common example would be a mixture of ethanoic acid and sodium ethanoate in solution. In this case, if the solution contained equal molar concentrations of both the acid and the salt, it would have a pH of 4.76. It wouldn't matter what the concentrations were, as long as they were the same.

You can change the pH of the buffer solution by changing the ratio of acid to salt, or by choosing a different acid and one of its salts.

Alkaline buffer solutions

An alkaline buffer solution has a pH greater than 7. Alkaline buffer solutions are commonly made from a weak base and its conjugate-acid or and one of its salts.

A frequently used example is a mixture of ammonia solution and ammonium chloride solution. If these were mixed in equal molar proportions, the solution would have a pH of 9.25. Again, it doesn't matter what concentrations you choose as long as they are the same.

How do buffer solutions work?

A buffer solution has to contain things which will remove any hydrogen ions or hydroxide ions that you might add to it - otherwise the pH will change. Acidic and alkaline buffer solutions achieve this in different ways.


Acidic buffer solutions

We'll take a mixture of ethanoic acid and sodium ethanoate as typical.

Ethanoic acid is a weak acid, and the position of this equilibrium will be well to the left:

Adding sodium ethanoate to this adds lots of extra ethanoate ions. According to Le Chatelier's Principle, that will tip the position of the equilibrium even further to the left.

The solution will therefore contain these important things:

• lots of un-ionised ethanoic acid; • lots of ethanoate ions from the sodium ethanoate; • enough hydrogen ions to make the solution acidic.

Other things (like water and sodium ions) which are present aren't important to the argument.

Adding an acid to this buffer solution

The buffer solution must remove most of the new hydrogen ions otherwise the pH would drop markedly.

Hydrogen ions combine with the ethanoate ions to make ethanoic acid. Although the reaction is reversible, since the ethanoic acid is a weak acid, most of the new hydrogen ions are removed in this way.

Since most of the new hydrogen ions are removed, the pH won't change very much - but because of the equilibria involved, it will fall a little bit.

Adding an alkali to this buffer solution

Alkaline solutions contain hydroxide ions and the buffer solution removes most of these.

This time the situation is a bit more complicated because there are two processes which can remove hydroxide ions.

Removal by reacting with ethanoic acid

The most likely acidic substance which a hydroxide ion is going to collide with is an ethanoic acid molecule. They will react to form ethanoate ions and water.

Because most of the new hydroxide ions are removed, the pH doesn't increase very much.


Removal of the hydroxide ions by reacting with hydrogen ions

Remember that there are some hydrogen ions present from the ionisation of the ethanoic acid.

Hydroxide ions can combine with these to make water. As soon as this happens, the equilibrium tips to replace them. This keeps on happening until most of the hydroxide ions are removed.

Again, because you have equilibria involved, not all of the hydroxide ions are removed - just most of them. The water formed re-ionises to a very small extent to give a few hydrogen ions and hydroxide ions.

Alkaline buffer solutions

We'll take a mixture of ammonia and ammonium chloride solutions as typical.

Ammonia is a weak base, and the position of this equilibrium will be well to the left:

Adding ammonium chloride to this adds lots of extra ammonium ions. According to Le Chatelier's Principle, that will tip the position of the equilibrium even further to the left.

The solution will therefore contain these important things:

• lots of unreacted ammonia; • lots of ammonium ions from the ammonium chloride; • enough hydroxide ions to make the solution alkaline.

Other things (like water and chloride ions) which are present aren't important to the argument.

Adding an acid to this buffer solution

There are two processes which can remove the hydrogen ions that you are adding.

Removal by reacting with ammonia


The most likely basic substance which a hydrogen ion is going to collide with is an ammonia molecule. They will react to form ammonium ions.

Most, but not all, of the hydrogen ions will be removed. The ammonium ion is weakly acidic, and so some of the hydrogen ions will be released again.

Removal of the hydrogen ions by reacting with hydroxide ions

Remember that there are some hydroxide ions present from the reaction between the ammonia and the water.

Hydrogen ions can combine with these hydroxide ions to make water. As soon as this happens, the equilibrium tips to replace the hydroxide ions. This keeps on happening until most of the hydrogen ions are removed.

Again, because you have equilibria involved, not all of the hydrogen ions are removed - just most of them.

Adding an alkali to this buffer solution

The hydroxide ions from the alkali are removed by a simple reaction with ammonium ions.

Because the ammonia formed is a weak base, it can react with the water - and so the reaction is slightly reversible. That means that, again, most (but not all) of the the hydroxide ions are removed from the solution.


(worksheet 2nd)




Calculate [H+] and [OH-] in buffer solutions

pH of buffer solutions can be determined by using acid-base indicators or by calculating of concentration of hudrogen ion or hydroxide ion.The components of buffer solutions are weak acid and its conjugate-base (acidic buffer) or weak base and its conjugate-acid (alkaline buffer), so what is the relationship between [H+] and the acid dissociation constant (Ka) or [OH-] and the base dissociation constant (Kb). In the next activity you will be understand the both relationships.. 1. The acidic buffer solutions

No Steps Solutions 1 The acidic buffer solution consists

weak acid HA and its salt LA. Write the ionization of each substance in water!

Ionizations : HA (aq) ↔ ….(aq) + …(aq) (1a) LA (aq)→ …..(aq) +… (aq) (1b)

2 Write the formulae of Ka of HA from reaction 1a

Ka = …………………… (2)

3 The adding of A- ion from salt, LA can shift the equilibrium system of HA to the left ( ractants), so [A-] from ionization of HA is decreasing. The number of [A-] from salt, LA in system are greater extremely than [A-] from HA . The [A-] in solutions that contribute in calculation are only the quantity of [A-] from salt. Write the relation between [H+], Ka, [HA] and [A-].

[H+] = ………………………..(3)

Where : Ka = The acid dissociation constant [HA] = concentration of weak acid [A-] = concentration of conjugate-base / anion of salt



2. The alkaline buffer solution No Steps Solutions 1 The acidic buffer solution consists

weak base MOH and its salt MA. Write the ionization of each substance in water!

Ionizations : MOH(aq) ↔ ….(aq) + …(aq) (1a) MA (aq)→ …..(aq) +… (aq) (1b)

2 Write the formulae of Kb of MOH from reaction 1a

Kb = …………………… (2)

3 The adding of M+ ion from salt, MA can shift the equilibrium system of MOH to the left ( ractants), so [M+] from ionization of MOH is decreasing. The number of [M+] from salt, MA in system are greater extremely than [M+] from MOH . The [M+] in solutions that contribute in calculation are only the quantity of [M+] from salt. Write the relation between [OH-], Kb, [MOH] and [M+].

[H+] = ………………………..(3)

Where : Kb = The base dissociation constant [MOH] = concentration of weak base [M+] = concentration of conjugate-acid / cation of salt

Conclusion

Notes


1. Write the formulae of [H+] of the following buffer solutions:

a. The mixture of HCOOH and HCOONa aqueous solution b. The mixture of HCN and Ca(CN)2 aqueous solution

2. Write the formulae of [H+] of the following buffer solutions:

a. The mixture of NH3 and NH4+ aqueous solution

b. The mixture of NH3 and (NH4)2SO4 aqueous solution

3. Calculate the pH of 600 mL of HCN 0.01 M dand 100 mL of KCN 0.01 M, given that Ka HF = 6 x 10-4

4. Calculate the pH of 100 mL of HF 0,06 M and 100 mL of BaF2 0.01 M, given that Ka HF = 6 x 10-4

5. Determine the pH of 180 mL of NH3 0.01 M and 100 mL ofNH4NO3 0.01 M , given that Kb NH3 = 1.8 x 10-5

6. Determine the pH of 200 mL of NH3 0.01 M and 90 mL of (NH4)2SO4 0.01 M, given that Kb NH3 = 1.8 x 10-5


7. What is the volume of acetic acid 0,01M must be added into 200 mL of sodium acetic 0.1M to form the buffer solution that has pH = 5.

Given that Ka CH3COOH = 1 x 10-5

8. How many grams of ammonium chloride (Mr = 53,5) must be added into 200 mL of ammonia 0,1M to form the buffer solution that has pH = 9.

Given thatKb NH3 = 1 x 10-5

9. Determine the proportion of volume of H2CO3 0.01 M and HCO3- 0.5 M to form the

buffer solution that has pH = 6 – log 5 (Ka H2CO3 = 5 x 10-7)

10. Calculate the pH of 200 mL larutan HF 0.1 M and 100 mL of Ba(OH)2 0.025 M. Given that Ka HF = 6 x 10-4

11. Calculate the pH of 200 mL of NH3 0.1 M and 100 mL of HCl 0.1M. Given that Kb NH3 = 1 x 10-5

12. How many grams of NaOH powder (Mr = 40) must be added into 500 mL of acetic acid 0.1 M to form the buffer that has pH = 5. Ka CH3COOH = 1 x 10-5


13. How many drops of calcium hydroxide 1 M must be dropped and dropped into 100 mL of acetic acid 0.01 M to form the buffer that has pH= 5 - log 9 . (Ka CH3COOH = 1 x 10-5)

14. How many dropps of sulfuric acid 1 M must be dropped and dropped intoharus 100 mL of ammonia 0.01 M to form the buffer that has pH-nya 9 + log 9 (Kb NH3 = 1 x 10-5)

15. One liter of the buffer solution that made from barium hydroxide aqueous solution 0.1 M and hydrofluoric acid 0.1 M has pH = 4 – log 6. When Ka HF = 6 x 10-4, determine the volume of each solution in the initial mixture.

16. Five hundreds milliliters of buffer solution that made from amonia 0.1 M and sulfuric acid 0.1 M has pH = 9. If Kb NH3 = 1 x 10-5, determine the volume of each solution in the initial mixture.

17. Five hundreds milliliters of buffer solution that made from amonia 0.1 M and sulfuric acid 0.1 M has pH = 9 - log 6. If Kb NH3 = 1 x 10-5,

a. determine the volume of each solution in the initial mixture. b. Determine the pH after the buffer is added by 50 mL of sulfuric acid 0.2 M


(worksheet 3rd)




Adding an acid and an alkali to buffer solutions

Purposes : Students are able to

1) explain the effect of adding a little acid to buffer solution 2) explain the effect of adding a little alkali to buffer solution

Material and aparattus

No Material /aparattus Size Quantity 1 Reaction tube - 11 / 1 2 Silinder ukur 10 mL 1 3 Pipette - 8 4 HCl (aq) 0.1 M 25 mL 5 NaOH (aq) 0.1 M 25 mL 6 CH3COOH (aq) 0.1 M 3 mL 7 CH3COONa (aq) 0.1 M 3 mL 8 NH3 (aq) 0.1 M 3 mL 9 NH4Cl (aq) 0.1 M 3 mL 10 Aquadest - - 11 Universal indicator - -

Procedure

PART I 1. Prepare 2 pieces of clean reaction tube 2. Pour to each tube above, 2 mL of hydrochloric acid 0.1 M and 2 dropps of

universal indicator, and measure its pH. 3. Do the following experiments :

a. Into the first tube (1.1), add dropp by dropp the solution of NaOH 0.1 M until the colour changes. Write the number of dropp and pH.

b. Into the second tube (1.2), dilute two times by adding 2 mL of aquadest, observe its colour!, if the colour has not changed, dilute 3 times by adding 2 mL of aquadest , etc until the colour changes. Write the volume of adding aquadest and its pH.

PART II 1. Prepare 2 pieces of clean reaction tube 2. Pour to each tube above, 2 mL of sodium hydroxide 0.1 M and 2 dropps of

universal indicator, and measure its pH. 3. Do the following experiments :

a. Into the first tube (1.1), add dropp by dropp the solution of HCl 0.1 M until the colour changes. Write the number of dropp and pH.



b. Into the second tube (1.2), dilute two times by adding 2 mL of aquadest, observe its colour!, if the colour has not changed, dilute 3 times by adding 2 mL of aquadest , etc until the colour changes. Write the volume of adding aquadest and its pH.

PART III 1. Prepare 3 pieces of clean reaction tube 2. Pour to each tube above, the mixture of 1 mL of acetic acid 0.1 M and 1 mL of

sodium acetic 0.1 M, then add 2 dropps of universal indicator, and measure its pH. 3. Do the following experiments :

a. Into the first tube (1.1), add dropp by dropp the solution of HCl 0.1 M until the colour changes. Write the number of dropp and pH.

b. Into the first tube (1.2), add dropp by dropp the solution of NaOH 0.1 M until the colour changes. Write the number of dropp and pH.

c. Into the second tube (1.3), dilute two times by adding 2 mL of aquadest, observe its colour!, if the colour has not changed, dilute 3 times by adding 2 mL of aquadest , etc until the colour changes. Write the volume of adding aquadest and its pH.

PART III 1. Prepare 3 pieces of clean reaction tube 2. Pour to each tube above, the mixture of 1 mL of ammonia 0.1 M and 1 mL of

ammonium chloride 0.1 M, then add 2 dropps of universal indicator, and measure its pH.

3. Do the following experiments : a. Into the first tube (1.1), add dropp by dropp the solution of HCl 0.1 M until

the colour changes. Write the number of dropp and pH. b. Into the first tube (1.2), add dropp by dropp the solution of NaOH 0.1 M

until the colour changes. Write the number of dropp and pH. c. Into the second tube (1.3), dilute two times by adding 2 mL of aquadest,

observe its colour!, if the colour has not changed, dilute 3 times by adding 2 mL of aquadest , etc until the colour changes. Write the volume of adding aquadest and its pH.

Observation result PART I : 2 mL of HCl 0.1 M is added by NaOH 0.1 M and diluted

colour pH change The solution that added

Volume dropps /

mL initial after initial after

NaOH 0.1 M

Aquadest

PART II : 2 mL of NaOH 0.1 M is added by HCl 0.1 M and diluted


Volume dropps /


HCl 0.1 M

Aquadest


PART III: 1 mL of CH3COOH 0.1 M and 1 mL of CH3COONa 0.1 M are added by HCl 0.1 M, NaOH 0.1 M and diluted


Volume dropps /


HCl 0.1 M

NaOH 0.1 M

Aquadest

PART IV: 1 mL of NH30.1 M and 1 mL of NH4Cl 0.1 M are added by HCl 0.1 M,

NaOH 0.1 M and diluted colour pH change The solution

that added Volume dropps /


HCl 0.1 M

NaOH 0.1 M

Aquadest

Assignments Based on the data above, plot the graph of volume versus the pH changes. ADDING AN ACID NaOH 0.1 M An acidic buffer solution An alkaline buffer solution pH

14 13 12 11 10 9 8 7 6 5 4 3 2 1

Volume of HCl 0.1 M Volume of HCl 0.1 M Volume of HCl 0.1 M


ADDING AN ALKALI HCl 0.1 M An acidic buffer solution An alkaline buffer solution pH

14 13 12 11 10 9 8 7 6 5 4 3 2 1

Volume of NaOH 0.1 M Volume of NaOH 0.1 M Volume of NaOH 0.1 M

DILUTION HCl 0.1 M NaOH 0.1 M Acidic buffer alkaline buffer pH

14 13 12 11 10 9 8 7 6 5 4 3 2 1

Volume of H2O Volume of H2O Volume of H2O Volume of H2O

Conclusion Based on the slope of the graph above , conclude the effect of adding a little acid, a little alkali and dilution to the change of pH of an acid, an alkali, an acidic buffer, and an alkaline buffer solution.


(worksheet 1st)


(Public Senior High School ) Jl. Sultan Agung Utara No. 7 Telp (0341) 324768 Malang 65111

(Web site : http://www.har-chemist.co.nr )

Before we study about hydrolysis of salt, we should review about definition of salt, the reaction of salt formation, the kind of salt based on the parent acid and parent base that react to produce salt. Discuss the following questions!

No Questions answer 1 Based on the chemical formula of salt,

mention the components of salt! ………………………………………. ……………………………………….

2 Based on the reaction of salt formation , what are the substances that react to produce the salt?

………………………………………. ……………………………………….

3 Based on the strength of acid and base that combine ( the parent acid and parent base), mention the kind of salt!

………………………………………. ………………………………………. ………………………………………. ……………………………………….

4 Give the example of salt, based on the question no 3!

………………………………………. ……………………………………….

MEASURING OF THE pH OF SALTS

PURPOSES Students are able to :

1) measure the pH of salts by experiments 2) conclude the relation between the pH of salt and the kind of salt based on the

strength of acid and base that combine



MATERIAL AND APARATTUS No Material / aparattus Measure Quantity 1 palette - 1 2 pipette - 6 3 NaCl (aq) 1 M 2 mL 4 Na2CO3(aq) 1 M 2 mL 5 CH3COONa(aq) 1 M 2 mL 6 NH4Cl(aq) 1 M 2 mL 7 Al2(SO4)3(aq) 1 M 2 mL 8 Ba(NO3)2(aq) 1 M 2 mL 9 Litmus ( red and blue) - 6 10 Universal indicator - 6

PROCEDURE 1. Drop the solution of each salt to palette about 3 dropps 2. Test by litmus and observe the colour change of litmus 3. Measure the pH of salt solution by universal indicator

OBSERVATION RESULT

THE COLOUR CHANGES OF nO SOLUTIONS RED LITMUS RED LITMUS

pH

ACIDIC / BASIC / NEUTRAL

1 NaCl (aq) ……………… ……………… ………… ……………… 2 Na2CO3(aq) ……………… ……………… ………… ……………… 3 CH3COONa(aq) ……………… ……………… ………… ……………… 4 NH4Cl(aq) ……………… ……………… ………… ……………… 5 Al2(SO4)3(aq) ……………… ……………… ………… ……………… 6 Ba(NO3)2(aq) ……………… ……………… ………… ………………

QUESTIONS

nO QUESTIONS ANSWER 1 Based on the experiments above, mention

the properties of each salt solution, acidic / alkaline or neutral

…………………………………… …………………………………… …………………………………… ……………………………………

2 If the salts that hydrolized in water are not the neutral salts, mention which are the salts that hydrolized in the experiment above!

…………………………………… …………………………………… …………………………………… ……………………………………


4. Complete the following table! The parent base The parent acid No Solution

Chemical formulae

Strong / weak

Chemical formulae

Strong / weak

1 NaCl (aq) 2 Na2CO3(aq) 3 CH3COONa(aq) 4 NH4Cl(aq) 5 Al2(SO4)3(aq) 6 Ba(NO3)2(aq)

CONCLUSION 1. The kind of salt based on the strength of parent acid and parent

base The strength of No

The parent base The parent acid The property of salt

( acidic / basic / neutral) 1 2 3 4 2. The kind of salts that hydrolized in water are : ……………………………………………………….. ……………………………………………………….. ………………………………………………………..

1. Complete the following table! The Parent Base of salt The Parent Acid of salt No Solution Chemical formulae

Strong / weak

Chemical formulae

Strong / weak

1 NaNO3 (aq) 2 Ba(CH3 COO)2 (aq) 3 (NH4)2SO4 (aq) 4 NH4CN(aq) 5 KI(aq) 6 CaS(aq)


2. The following salts, hydrolized or not in water? Explain! a. Sr(ClO4)2 b. NaF c. NH4Br d. CH3COONH4


Hydrolysis Reactions A salt of a strong acid and a strong base (such as NaCl from HCl and NaOH)

produces a neutral solution when dissolved in water. However, when a salt of a weak acid

and a strong base (e.g. NaAc from acetic acid and NaOH), a strong acid and a weak base

(e.g. NH4Cl from ammonia and HCl), or a weak acid and a weak base (NH4Ac) is

dissolved in water, the solution does not have a neutral pH. These phenomena are

explained by the reaction of the salts of weak acids and weak bases with water in

hydrolysis reactions. As shown in Figure 2.1, these hydrolysis reactions produce the

parent weak acids and weak bases of the salts:

Figure 2.1: Hydrolysis reactions

As you can see in the figure above, the salt of a weak acid, such as acetate ion, acts

as a base in water, and the salt of a weak base, such as ammonium ion, acts as a weak

acid. From our previous discussion on the reactions of acids and bases with water in

Disassociations, you should know that the Kb of acetate ion can be calculated from the Ka

of acetic acid, and that the Ka of ammonium ion can be calculated from the Kb of

ammonia, as shown in figure 2.1.

Another type of hydrolysis reaction comes from the reaction of metal ions with

high charges. Such ions act as Lewis acids to water molecules, as shown in Figure 2.2. A

metal ion can bond to water by accepting an lone pair from the oxygen of a water

molecule, and this increases the acidity of water molecule. Like other acids, we can

calculate the Ka and calculate the pH of a solution containing such ions.

Figure 2.2: Why metal ions decrease the pH of aqueous solutions

To calculate the pH of a solution containing the salt of a weak acid or a weak base,

treat the problem exactly as you did when calculating the pH of weak acid and base

solutions above in Calculating pH's, Heading . Mixtures of salts of weak acids and weak

bases present a challenging mathematical problem that we will not cover in our treatment

of acid-base chemistry.


http://www.sparknotes.com/chemistry/acidsbases/fundamentals/section1.html

http://www.sparknotes.com/chemistry/acidsbases/phcalc/section1.html

Terms Acid - A substance that has the potential to donate a proton or accept an electron pair.

Acidic - Having a pH less than 7 or a pOH greater than 7.

Base - A substance that can accept a proton, release OH-, or donate an electron pair.

Basic - Having a pH greater than 7 or a pOH less than 7.

Hydrolysis - A reaction that modifies a water molecule. In acid-base chemistry this

usually refers to the reaction of a solute which changes the pH of an aqueous solution.

Parent acid - An acid that reacts with base to produce the salt

Parent base - A base that reacts with acid to produce the salt

pH - A measure of the hydrogen ion concentration, it is equal to - log [H+].

pKa - A measure of the strength of an acid, it is equal to - log Ka, where Ka is the acid

dissociation constant in water.

pKb - A measure of the strength of a base, it is equal to log Kb, where Kb is the base

dissociation constant in water.

pOH - A measure of the hydroxide ion concentration, it is equal to - log [OH- ].

Strong Acid - An acid with a pKa less than zero. Strong acids completely dissociate in

water.

Strong Base - A base with a pKb less than zero. Strong bases completely dissociate in

water.

Weak Acid - An acid with a pKa greater than zero. Weak acids do not completely

dissociate in water.

Weak Base - A basewith a pKb greater than zero. Weak bases do not completely

dissociate in water.

Weakness of attitude become weakness of character (Albert Einstein)


(worksheet 2nd)


(Public Senior High School ) Jl. Sultan Agung Utara No. 7 Telp (0341) 324768 Malang 65111

(Web site : http://www.har-chemist.co.nr )

Determaining [OH-] or [H+] of hydrolyzed salt solutions The pH of salt solutions can be measured by universal indicator, also can be determained by calculation. In the previous activity, we have concluded that the salts from strong base and acid are not hydrolyzed, so their pH equal with water’s pH (neutral). But the other salts ( from weak base-strong acid or strong base-weak acid) are hydrolyzed, so their property are acidic, or basic. How to calculate the pH of salt solution? In the next activity, you will determain the relation among hydrolysis constant (Kh), dissociation constant of water (Kw), [OH-] or [H+] in the salt solutions 1. THE SALT FROM THE STRONG BASE (LOH) AND STRONG ACID (HZ)

No Questions Solution 1 The chemical formulae of salt LZ

2 Ionization in water LZ → ......... + ............

(cation of strong base) (anion of strong acid)

3 Hydrolysis in water Not hydrolyzed

4 Hydrolysis constant (Kh) Doesn’t have Kh

5 The value of [H+] [H+] = [H+] in water = Kw

2. THE SALT FROM STRONG BASE (LOH) AND WEAK ACID (HA)

No Questions Solution 1 The chemical formulae of salt LA (1)

2 Ionization in water LA → ......... + ............ (2)

(cation of strong base) (anion of weak acid) 3 Hydrolysis in water

(the weak ion) A- (aq)+H2O (l) ↔HA(aq) + OH-(aq) (3) Produced OH- ion in solution causes the property of salt :............ (acidic/basic/neutral)

4 The hydrolysis constant (Kh) a. The hydrolysis constant of A- ion (

reaction 2)

a. The hydrolysis constant of A- ion K = ............................. (4a)



b. From the formulae of Ka (question 4a), the value of [H2O] doesn’t change, so K.[H2O] can be expressed by Kh

b. If K.[H2O] is expressed by Kh, so : Kh = ............................. (4b)

5 The relation among Kh, Kw and Ka a. Write the formulae of Kw b. Write ionization of weak acid, HA

and write theformulae of its Ka

c. The formulae of Kh (4b), multiply by :

][][

+

+

HH

d. Substitute Kw and 1/Ka to

equation (5c) above, so you will get the formulae of Kh of salt from strong base and weak acid.

a. Ionization of water : H2O ↔ ........ + ........ Kw = ........................ (5a)

b. HA ↔ ........ + .......... Ka =...........................

=Ka1

............................. (5b)

c. Kh = .......................... (5c)

d. Kh = ............................ (5d)

5 The value of [OH-] Please observe the hydrolysis of A – ion that [HA] = [OH-], so the multiplication of [HA]. [OH-] expressed by [OH-]2. Substitute this equation the formulae of Kh (equation 4b)

Kh =........................ [OH-]2 = ................... [OH-] = ............ (6)

3. THE SALT FROM WEAK BASE (MOH) AND STRONG ACID (HZ)

No Pertanyaan Penyelesaian 1 The chemical formulae of salt MZ (1)

2 Ionization in water MZ → ......... + ............ (2)

(Cation of weak base (anion of strong acid) 3 Hydrolysis in water

(the weak ion) M+ (aq)+H2O (l) ↔MOH(aq) + H+(aq) (3) Produced H+ ion in solution causes the property of salt :............ (acidic/basic/neutral)

4 The hydrolysis constant (Kh) c. The hydrolysis constant of A- ion (

reaction 2) d. From the formulae of Ka

(question 4a), the value of [H2O] doesn’t change, so K.[H2O] can be expressed by Kh

a. The hydrolysis of constant of M+ ion

K = ............................. (4a)

b. If K.[H2O] expressed by Kh, so : Kh = ............................. (4b)

5 The relation among Kh, Kw and Ka a. Write the formulae of Kw b. Write ionization of weak acid, HA

and write theformulae of its Ka

a. Ionization of water : H2O ↔ ........ + ........ Kw = ........................ (5a)

b. MOH ↔ ........ + .......... Kb =...........................


c. The formulae of Kh (4b), multiply by :

][][

−

−

OHOH

d. Substitute Kw and 1/Kb to equation (5c) above, so you will get the formulae of Kh of salt from weak base and strong acid.

=Kb1

............................. (5b)

c. Kh = .......................... (5c)

d. Kh = ............................ (5d)

6 The value of [OH-] Please observe the hydrolysis of M+ ion that [MOH] = [H+], so the multiplication of [MOH]. [H+] expressed by [H+]2. Substitute this equation the formulae of Kh (equation 4b)

Kh =........................ [H+]2 = ................... [H+] = ............ (6)

4. THE SALT FROM WEAK BASE (MOH) AND WEAK ACID (HA)

No Questions Solution 1 The chemical formulae of salt MA (1)

2 Ionization in water MA → ......... + ............ (2)

(cation of weak base) (anion of weak acid) 3 Hydrolysis in water

(the weak ion ) M+ (aq)+H2O (l) ↔MOH(aq) + H+(aq) A- (aq)+H2O (l) ↔HA(aq) + OH-(aq) + M+ + A- +H2O ↔ MOH + HA (3)

4 The hydrolysis constant (Kh) a. Write the equilibrium constant

ionization of M+ (reaction 2) dan b. From the formulae of K

question 4a, the value of [H2O] doesn’t change, so K.[H2O] can be expressed by Kh

a. The equilibrium constant of reaction (3)

K = ............................. (4a)

b. If K.[H2O] can be expressed by Kh, so : Kh = ............................. (4b)

5 The relation among Kh, Kw and Kb a. Write the formulae of Kw, 1/Ka,

dan 1/Kb b. The formulae of Kh (4b), multiply

by

]].[[]].[[

−+

−+

OHHOHH

c. Substitute Kw ,1/Kb, and 1/Ka to equation (5b) above, so you will get the formulae of of salt from weak base and weak acid.

a. Kw = ........................ (5a1)

=Kb1

............................. (5a2)

=Ka1

........................... (5a3)

b. Kh = .......................... (5b)

c. Kh = ............................ (5c)

6 The value of[H+] Observe the hydrolysis of A- ion and M+ ion that [MOH] = [HA], and [A-] = [M+],

Kh = ]].[[]].[[

−+ AMHAMOH


so [MOH]. [HA]= [HA]2 and [M+].[A-] = [A-]2 Substitute this equation to the formula of Kh (equation 4b) Then multiply by:

2

2

][][

+

+

HH

= 2

2

][][

−AHA

x 2

2

][][

+

+

HH

Kh = 22 ].[1 +H

Ka

[H+] = ............ (6)

Questions

No Questions Answer 1 Based on the the strength of parent base and

parent acid, clasify the kind of salt and give 1 example of each salt!

.................................................................

.................................................................

.................................................................

2 Write the hydrolysis reaction of salt in number 1.

.................................................................

.................................................................

.................................................................

3 Based on the kind of cation and anion that hydrolyzed, mention the kind of hydrolysis .

.................................................................

.................................................................

.................................................................

Conclusion

Salts Partial

hydrolyzed/ Total hydrolyzed/

not hydrolyzed

Kh [H+] or

[OH-]

Property (acidic/basic/

neutral)

Notes

SB-SA Not hydrolyzed

- [H+]= Kw

neutral

SB-WA

[A-] = concentration anion of salt

WB-SA

[M+] = concentration of cation of salt

WB-WA

SB = strong base SA = strong acid WB = weak base WS = weak acid


1. The solution of Fe2(SO4)3 . a. Mention the parent base and acid b. Write the hydrolysis reaction c. What is the property of salt (acidic/basic/neutral)

2. Calculate the pH of the following salts : a. 200 mL of (CH3COO)2 Ca 0.05 M (Kh NaAc = 10- 9) b. 200 mL of (NH4)2 SO4 0.0005 M (Kb NH3 = 10- 5) c. 200 mL CH3COONH4 0,173M (Kb NH3 = 10- 5 and Ka CH3COOH = 10- 5)

3. The mixture of 50 mL of NH4OH 0.1 M + 50 mL of HCl 0.1 M ,Calculate the pH of before and after mixed. Kb amonia = 10- 5

4. The mixture 50 mL of Ba(OH)2 0.1 M + 100 mL of HCN 0.1M,Calculate

the pH of before and after mixed . Ka HCN = 10- 7 5. The mixture 50 mL of H2SO4 0.1 M + 100 mL of NH3 0.1 M ,calculate the pH before and after mixed. Kb NH3 = 10- 5

PREDICTION IS VERY DIFFICULT, ESPECIALLY ABOUT THE FUTURE (NIELS BOHR)


Worksheet 1st

Drs. M. Hariyanto

SMA Negeri 3 Malang (Public Senior High School)

Jl. Sultan Agung Utara No 7 Phone (0341) 324 768 Malang 65111 (web site : http://www.har-chemist.co.nr )

Although all compounds have a characteristic solubility in water at a given temperature, some families of compounds are more soluble than others and it is useful to know certain general rules of solubility. We call any substance the solubility of which is less than 0.01 mol/L insoluble. If its solubility is greater than 0.1 mol/L, we call it soluble. If its solubility is between 0.01 and 0.1 mol/L, we say that it is slightly soluble. The following solubility rules can be used to determine solubilities in water, with the disclaimer that they don't always hold, nor do they include every ion which is in common use, but they are good to have nearby when needed: 1. All sodium, potassium, and ammonium salts are soluble. 2. All nitrates, acetates and perchlorates are soluble. 3. All silver, lead and mercury(I) salts are insoluble. 4. All chlorides, bromides and iodides are soluble. 5. All carbonates, sulfides, oxides and hydroxides are insoluble. 6. All sulfates are soluble except calcium sulfate and barium sulfate.

A. Solubility 1. Definition of solubilty

Complete the following table!

The amount of substance in water

The amount of dissolved in

water

Volume of solvent ( dm3)

solubility

15 grams of P 3 grams 10 0,3 gram/dm3 25 moles of Q 0,2 moles 20 ................. 10 grams of R 0,04 mgrams 1 ................ 5 moles of S 0,01 moles 4 ................



Questions Based on the data above, answer the following questions!

No Questions answer 1 What is solubility?

...........................................................

2 What is the unit of solubility?

...........................................................

3 Write the equation of relationship among solubility (s), number of mole of solute (n), and volume of solvent (V)

...........................................................

...........................................................

2. The factors that influence the value of solubility of substance in water What are the factors that influence the value of solubility? Observe the following data, and answer the questions!

The solubility at The solubility in solvent The amount of substance

25oC 50oC water Alcohol 10 grams of A 0,1 g/dm3 0,2 g/dm3 0,15 g/dm3 0,1 g/dm3 10 grams of B 0,05 g / dm3 0,075 g/dm3 0,2 g/dm3 0,3 g/dm3

Questions Based on the data above, answer the following questions!

No Questions Answer 1 Based on the data above, what are

the factors that influence the value of solubility of substance in water?

..........................................................

..........................................................

..........................................................

1. If the solubility of BaSO4 in water is 1 x 10-5 mol/L, determine the solubility of BaSO4 by

unit mg/L (Mr BaSO4 = 233) 2. The solubility of Mg(OH)2 in water is 2 x 10-4 M, how many grams of Mg(OH)2 that can

soluble in 500 mL of water ( Mr Mg(OH)2=58) 3. The maximum mass of L2X3 (Mr=100) that can dissolve in 200 mL of water is 5 x 10-3 mg.

Calculate the solubility of L2X3 in water by unit moles/L


B. Solubility Product Constant (Ksp)

Expression of Ksp

Before you write the expression of Ksp (solubility product constant) of saturated substance, answer the following questions!

No Questions Answer 1 Ag2CrO4 is the saturated substance,

Write the ionization of Ag2CrO4 in water!

Ag2CrO4(s) ↔ .......(aq) + ........(aq)

2 Write the equilibrium constant (K) of equation no 1

K = ................................................

3 If the equilibrium constant of saturated substance is expressed by Ksp, write the Ksp of Ag2CrO4!

Ksp Ag2CrO4 = .................................

Individual assignments

Fill the blank ! NO The chemical

formulae The equilibrium equation The formulae of

Ksp 1 AgCl AgCl(s) ↔ Ag+

(aq) + Cl-(aq) Ksp=[Ag+][Cl-]

2 CaCO3 3 Ag2S 4 Fe(OH)3 5 CuCl2 6 Al2(SO4)3 7 PbCl2 8 BaSO4 9 Al(OH)3

10 BaF2

Conclusion If the saturated substance is symbolized by LxAy, Write the ionization equation and its Ksp expression


The relation between Ksp and the solubility of

substance in water (s)

Individual assignments No Questions Answer 1 If the saturated substance is symbolized

by LxAy, Write the ionization equation

LxAy(s) ↔ ........(aq) + .......(aq)

2 If the concentration of LxAy equal with its solubility (= s), determine the [Ly+] and [Ax-] by using coefficient ratio!

LxAy(s) ↔ ........(aq) + .......(aq) s ~ ........... ~ .........

3 Write the formulae of Ksp of LxAy, and substitute the value of [Ly+] and[Ax-] from equation no 2, so you get the relationship between Ksp and s. Where: s = solubility of substance in water; x = the coefficient of cation; y = the coefficient of naion

Ksp LxAy = ........................ or s = .............................

1. Write the relationship between Ksp and s.

NO The chemical

formulae The formulae of Ksp The formulae of (S)

1 AgCl Ksp=[Ag+][Cl-] = s2 s = √Ksp 2 CaCO3 3 Ag2S 4 Fe(OH)3 5 CuCl2 6 Al2(SO4)3 7 PbCl2 8 BaSO4 9 Al(OH)3

10 BaF2 2.Calculate the Ksp of CuCl2 , if its solubility in water is 10-3M


3. Calculate the solubility of Ag2CrO4(s) , given that Ksp = 2,4 .10-12

4. Determine the Ksp of AgCl, if the solubility of AgCl in water = 0,7175 mg/500 mL 5. How many miligrams of Ag3PO4 (Mr=419) that can soluble in 500 mL of water, given that

its Ksp = 2,7 x 10-11?


Worksheet 2nd

Drs. M. Hariyanto

SMA Negeri 3 Malang (Public Senior High School)

Jl. Sultan Agung Utara No 7 Phone (0341) 324 768 Malang 65111 (web site : http://www.har-chemist.co.nr )

Precipitation reaction

When we mix two electrolytes substances that by theorytically produce precipitation, but in facts we will get 3 possibility of products; no precipitation occurs (unsaturated), will have precipitated (saturated) or produce precipitation (supersaturated) Why these are happened? Observe the following data! And complete it!

No Substance The concentration of ions in solution

The product of concentration of

ions or Trial product (Qc)

Ksp Precipitation

[Ag+] = 10-6 M [Br-] = 10-7 M

[Ag+].[Br-]= 10-13 AgBr has not precipitated (unsaturated)

Ag+] = 10-5 M [Br-] = 10-7 M

[Ag+].[Br-]= 10-12 AgBr has precipitated (supersaturated)

1 AgBr

Ag+] = 2. 10-6 M [Br-] = 2. 10-7 M

[Ag+].[Br-]= 4. 10-13

4 x 10-13

AgBr will have precipitated (saturated)

[Mg2+] = 10-6 M [OH-] = 10-4 M

............................ ............................

[Mg2+] = 10-5 M [OH-] = 10-2 M

............................ ............................

2 Mg(OH)2

[Mg2+] = 2.10-6 M [OH-] = 2.10-3 M

.............................

8 x 10-12

.............................

Conclusion

Based on the data , conclude the relation among Qc, Ksp and the forming of precipitation!



Example

1. Ksp for PbCl2 = 6,25 .10– 5, should precipitation occurs if 100 mL of Pb(NO3)2 0.005 M is mixed with 100 mL of HCl 0.005 M

Solution :

The number of mole of Pb(NO3)2 = 100 x 0.005 = 5 mmoles The number of mole of HCl = 100 . 0,005 = 5 mmoles PbCl2 → Pb2+ + 2 NO3

- 5 mmoles ∼ 5 mmoles ∼ 10 mmoles HCl → H+ + Cl- 5 mmoles ∼ 5 mmoles ∼ 5 mmoles Concentration of each ion after mixed. [Pb2+] = 5 mmoles /200 ml = 2.5 x 10-2 M [Cl- ] = 5 mmoles /200 ml = 2.5 x10-2 M The trial product or Qc of PbCl2 = [Pb2+] [Cl- ]2 =(2.5 x10-2)( 2.5 x10-2)2 = ………… Qc (> / < / = ) Ksp , so ……………………………..

1. One hundred mililitres of the saturated solution of BaF2 , contains 17.5 mg of precipitation after evaporated. Determine the Ksp for BaF2 , given that the relative atomic mass of Ba and F are 137 and 19.

2.How many grams of Al2(SO4 )3 that can dissolve in 250 mL of solution , if Ksp for

Al2(SO4)3 = 3,2 . 10-34.(Ar Al = 27; O = 16 ; S = 32.) 3. Calculate the pH of the saturated solution of Al(OH)3 , if Ksp for Al(OH)3 = 1,6. 10-35.


4.The pH of saturated solution of Zn(OH)2 is 8, determine Ksp for Zn(OH)2. 5. Should precipitation occurs of BaSO4 ,If 200 mL of Ba(NO3)2 0.025 M is mixed with 300

mL of Na2SO4 0.05 M. ( Ksp for BaSO4 = 1.5 x 10-9) 6.The solution of FeBr3 0.2 M is dropped by NH3 solution, determine the pH when start

precipitation of Fe(OH)3, Ksp for Fe(OH)3 = 1.6 x 10-36 ( assume that the volume of mixture doesn’t change by ammonia addition)

7. Ksp forBaSO4 = 1 x 10-10. If barium hydroxide that its pH= 11 is mixed with sulfuric acid

that its pH = 3 so the volume of mixture 1 Litre. a. predict ! Should precipitation occurs? b. if precipitation occurs, calculate the mass of BaSO4 that precipited (Mr=233) c. calculate the concentration of Ba2+ and SO4

2- in solution!. 8. One litre of MgCl2 0.0001 M is added drop by drop with KOH 2 M, if Ksp for

Mg(OH)2 = 9 x 10-12, how many drops of KOH required to start precipitation of Mg(OH)2 ?

9. Gived that Ksp for BaSO4 = 1 x 10-10, Ksp for MgSO4 = 1 x 10-5, Ksp for SrSO4 = 3 x 10-7

and Ksp for CaSO4 = 3 x 10-6. If into the 500 mL of the mixture of BaCl2, MgCl2, SrCl2 andCaCl2 that their concentration are 0.0002 M are added by 500 mL of Na2SO4 0.00002 M, determine which is the salt that precipited?


Solubility and the common ion effect

As in systems involving dissociation of weak electrolytes, the common ion effect is of importance in solubility equilibria. Le Chatelier’s principle prdicts a decrease in solubility of saturated substance if either a soluble common ion is added.

To review the effect of adding substance into the equilibrium systems, answer the following questions!

No Questions Answer 1 Given that the equilibrium in

saturated solution: Ag2SO4(s) ↔ 2Ag+(ag) + SO4

2-(aq) Where will the eqilibrium system shift? If :

a. added by SO42- ion

b. added by Ag+ ion

........................................................

........................................................

2 Determine the amount of Ag+ and SO4

2- if the equilibrium shifts to reactant!

........................................................

........................................................

Individual assignments

What will happen if AgCl is dissolved in NaCl solution or AgNO3 solution? Compare the solubility of AgCl in NaCl solution or AgNO3 solution with solubility of AgCl in water? Change or not?

No Questions answer 1 If AgCl is dissolved in NaCl x M

so it will happen two ionizations. Write the equation of the both ionizations!

AgCl(s) ↔ ......(ag) + .....(aq) NaCl(aq)→ ......(aq) + ......(aq)

2 Which is the ion of the product NaCl ionization that influences the equilibrium of AgCl? Where will the equilibrium shift?

........................................................

........................................................

3 Based on the direction of shifting no 2, What is the effect to the solubility of AgCl? Compare with the solubility of AgCl in water!

........................................................

........................................................

Conclusion


Example Given Ksp for BaSO4 = 1 x 10-10, Determine the solubility of BaSO4 in Na2SO4 0,01M. Solution : The solubility of BaSO4 in Na2SO4 0,01 M = x mol/L BaSO4(s) ↔ Ba2+ (aq) + SO4

2-(aq) x M ~ x M ~ x M Na2SO4(aq) → 2 Na+ (aq) + SO4

2-(aq) 0.01 M ~ 0.02 M ~ 0.01 M BaSO4(s) ↔ Ba2+ (aq) + SO4

2-(aq) Initial : a M 0 M 0.01 M Solubility : x M - ~ x M ~ x M + Equilibrium : (a-x) M x M (0.01 + x)M Concentration of ions at equilibrium : [Ba2+] = x M [SO4

2-] = ( 0.01 + x ) M ~ 0,01 M Note : x is very small compared to 0.01 and hence the approximation that ( 0.01 + x ) M = 0,01 M is justified. Ksp BaSO4 = [Ba2+][SO4

2-] ............... = ........ ........ x = ................. so the solubility of BaSO4 in Na2SO4 0.01 M is ..............................moles.L-1

1. Given that Ksp for AgCl = 1 x 10-10, determine the solubility of AgCl in:

a. water b. Na2SO4 0.1 M c. NaCl 0.01 M d. CaCl2 0.005 M e. AgNO3 0.01 M


2. The solubility of Mg(OH)2 in water is 2 x 10-4M. Determine the solubility of Mg(OH)2 in : a. Mg(NO3)2 0.01 M b. NaOH 0.0001 M c. The solution that has pH = 10

3. The solubility of AgX in AgNO3 0.001 M is 4 x 10-5M. Determine : a. Ksp for AgX b. The solubility of AgX in water

Intellectuals solve problems; geniuses prevent them (Albert Einstein)


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