Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base

Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture.

Strong Acid–Strong Base

Mixture CalculationsExample 2

We’re given that 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.

125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.

And we’re asked to determine the pH of the final mixture.

125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

Just a few words about sulphuric acid, H2SO4.


H2SO4.


H2SO4

is a Diprotic Acid


H2SO4 is a Diprotic Acid

Which means it has 2 protons it can lose.


H2SO4 is a Diprotic AcidIt has 2 protons it can lose.

As soon as H2SO4 is added to water, it ionizes completely to lose its first proton:


When H2SO4 is added to water, it ionizes completely to lose its first proton:

100% of the H2SO4 molecules lose one proton (click) to form hydronium and hydrogen sulphate ions.



2 4 2 3 4H SO H O H O HSO 100% ionization

H+

But when its just in water, the second proton does not come off as easily. This proton comes off when HSO4 minus ionizes. But HSO4- is a weak acid so its ionization in water is very limited.



2 4 2 3 4H SO H O H O HSO 100% ionization

But its second proton does not come off as easily in water:

24 2 3 4HSO H O H O SO 100% ionization

EquilibriumWeak Acid

However, when H2SO4 is mixed with the STRONG BASE KOH, this is a totally different situation.


When H2SO4 is mixed with thestrong base KOH, this is a totally different situation.

When an H2SO4 molecule enters water, it loses one proton (click) to water, to form a hydronium ion (H3O+) and a hydrogen sulphate ion (HSO4 minus).


2 2 434 HH S O H OO OH S H+

Models of these are shown here. Take a moment to check the atoms and the charges and see how the formulas relate to the structural models.


OH

H

H +

O

O H

SO

O–

2 2 434 HH S O H OO OH S H+

When the strong base KOH dissociates in water it forms K+ and OH minus ions. Here we doubled everything in the equation.


2KOH 2K+ + 2OH–

OH

H

H +

O

O H

SO

O–

we show models of the two hydroxide ions from the KOH


OH

HO

–

–

2KOH 2K+ + 2OH–

OH

H

H +

O

O H

SO

O–

One of the hydroxide ions collides with the hydronium ion


HO

OH

–

–

OH

H

H +

O

O H

SO

O–

and takes away a proton, to form 2 water molecules


HO–

OH

H

H OH

O

O H

SO

O–

The other hydroxide ion collides with the hydrogen sulphate ion (click)


HO

H OH

–

OH

H

O

O H

SO

O–

And takes a proton from it to form a water molecule and a suphate ion


O

OS

O

O–

H OH

HO

H

–

OH

H

the sulphate ion has the formula SO4 2 minus.


H OH

HO

H

O

OS

O

O–

–

Sulphate SO4

2–

OH

H

So, in an indirect way, 2 hydroxide ions are able to remove both protons from a molecule of H2SO4. We‘ll show this with equations.


Two OH– ions were able to remove both protons from H2SO4.

As soon as H2SO4 is added to water it ionizes completely to form a hydronium ion and a hydrogen sulphate ion. We’ll call this Step 1



2 4 2 3 4H SO H O H O HSO Step 1

When we add a strong base, one OH minus ion neutralizes the hydronium ion to form 2 water molecules. We’ll call this step 2



2 2 434 HH SO O H OOH S

OH+

22H O

Step 2

And the other OH minus ion reacts with hydrogen sulphate to form water and a sulphate ion. We’ll call this Step 3.



42 4 2 3H SO H O H HO SO

OH+

22H O

OH+

22 4H O SO

Step 3

Even though we know these 3 steps occur when we add H2SO4 to water and then add a strong base,


Even though we know that these steps occur…

1

2 3

We can represent the process with a net overall equation: H2SO4 plus 2 OH minus form 2H2O plus SO4 2minus.


Even though we know that these steps occur…

22 42 4H SO 2OH 2H O SO

We can represent the process with a net overall equation.

1

2 3

so in the overall net reaction, we see that each H2SO4 (click) donates 2 protons or H+ ions to the hydroxide ions.


422

2 4SO 2OHH 2H O SO 2 H+

Each H2SO4 donates

2 protons to the OH– ions

From this, we can write the conversion factor stating there are 2 moles of H+ per 1 mole of H2SO4. We can use this conversion factor in any calculation where H2SO4 reacts with a strong base.


422

2 4SO 2OHH 2H O SO

2 H+

Each H2SO4 donates

2 protons to the OH– ions 2 4

2 mol H

1 mol H SO

Now we’ll do the calculations for this problem. We’ll begin by calculating the initial moles of H+ added.


2 4

2 4

0.150mol H SO 2momol

l H0.125L 0.0375mol H

1L 1mol H SOHinitial

0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH

1L 1mol KOHinitial

+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H

+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L

3pH log H O log 0.0273 1.56 2 4

2 mol H

1 mol H SO

Its equal to 0.150 moles of H2SO4 per Litre…


2

2 4

4 2mol Hmol H 0.125L 0.0375mol H

1

0.150

mol H SO

mol H SO

1Linitial


1L 1mol KOHinitial


+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L

3pH log H O log 0.0273 1.56 2 4

2 mol H

1 mol H SO

Times 2 moles of H+ to 1 mole of H2SO4…

2

2 4

4 2mol H

1mo

0.150mol H SOmol H 0.125L 0.0375mol H

1 l H SL Oinitial


1L 1mol KOHinitial


+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L

3pH log H O log 0.0273 1.56 2 4

2 mol H

1 mol H SO


Times 0.125 L

2 4

2 4

0.150mol H SO 2mol Hmol H 0.0375mol0 H

1L 1mol SO5L

H.12initial


1L 1mol KOHinitial


+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L

3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

Which comes to 0.0375 moles of H+. Notice moles of H2SO4 and Litres cancel out.

2 4

2 4

0.150mol H SO 2mol Hmol 0.0375molH 0.125L



1L 1mol KOHinitial


+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L


In order to preserve 3 significant figures (the lowest number of significant figures in the given data)…

2 4

2 4

0.150mol H SO 2mol Hmol H 0.125L 0.0 mol37 H

1L 1mol H SO5initial


1L 1mol KOHinitial


+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L


3 significant

figures

The answer to this must be expressed to 4 decimal places.

2 4

2 4

0.150mol H SO 2mol Hmol H 0.125L 0. mol037 H

1L 1mol H S5

Oinitial


1L 1mol KOHinitial


+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L


4 decimal places

Now we’ll calculate the initial moles of OH minus added.

2 4

2 4

0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H

1L 1mol H SOinitial

0.200mol KOH 1mol OH0.150L 0.0300mol OH

1L 1molmol

KH

OHO initial


+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L


It is equal to 0.200 moles of KOH per L

2 4

2 4


1L 1mol H SOinitial

1mol OHmol OH 0.150L 0

0.2.0300mol OH

1mol KO

00mo

H

l KOH

1Linitial


+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L


Times 1 mole of OH minus to 1 mole of KOH

2 4

2 4


1L 1mol H SOinitial

1mol OH0.200mol KOHmo

1mol Kl OH 0.150L 0.0300mol OH

OH1Linitial


+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L


Times 0.150 L

2 4

2 4


1L 1mol H SOinitial

0.200mol KOH 1mol OHmol OH 0.0300mol OH

1L 1mol0.150L

KOHinitial


+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L


Which comes out to 0.0300 moles of OH minus. You can see that moles of KOH and Litres both cancel. Notice we also have 3 significant figures and 4 decimal places in this answer.

2 4

2 4


1L 1mol H SOinitial

0.200mol KOH 1mol OHmol OH 0.150L

1L 1mol0.0300mol

KOHOHinitial


+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L


Now we compare the initial moles of H+ and OH minus. We see that 0.0375, the moles of H+, is greater than 0.0300, the moles of OH minus

2 4

2 4




1L 1mol0.0300mol

KOHOHinitial


+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L


So the H+ is in excess and the OH minus is the limiting reagent.

2 4

2 4




1L 1mol0.0300mol

KOHOHinitial


+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L


In Excess

Limiting Reagent

We calculate the excess moles of H+…

2 4

2 4


1L 1mol H SOinitial


1L 1mol KOHinitial

+ 0.0375mol H 0.0300mol OHExce 0.0075mol Hssmol H

+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L


By taking 0.0375 moles of H+

2 4

2 4



+Excessmol H 0.0300mol OH0.0375mo 0.0075molH Hl

+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L



1L 1mol KOHinitial

and subtracting 0.0300 moles of OH minus.

2 4

2 4


1L 1mol H SOinitial

+ 0.Exc 030essmol 0mol OHH 0.0375mol H 0.0075mol H

+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L



1L 1mol0.0300mol

KOHOHinitial

To give us 0.0075 moles of H+ in excess

2 4

2 4


1L 1mol H SOinitial

+Excessmol H 0.0375mol H 0.0300mol 0.0075mol HOH

+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L



1L 1mol KOHinitial

This answer, 0.0075, has 4 decimal places, because the numbers we subtracted both had 4 decimal places.

2 4

2 4


1L 1mol H SOinitial

+Excessmol H 0. mol H 0. mol O0375 H 0. mol030 50 H007

+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L



1L 1mol KOHinitial

4 decimal places

4 decimal places

4 decimal places

But we can see that written this way, this number has only 2 significant figures, the 7 and the 5. Therefore the final answer to this problem cannot have more than 2 significant figures.

2 4

2 4


1L 1mol H SOinitial

+Excessmol H 0.0375mol H 0.0300mol OH 0.00 mol5 H7

+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L



1L 1mol KOHinitial

2 significant figures

The next step on the way to pH, is to find the hydronium ion concentration.

2 4

2 4


1L 1mol H SOinitial


+ +

3

0.0075mol H 0.0075mol H0.0273M

0.125L 0.150L 0.275LH O H



1L 1mol KOHinitial

Which is equal to the concentration of H+. These are synonymous in chemistry dealing with aqueous solutions.

2 4

2 4


1L 1mol H SOinitial


+ +

3

0.0075mol H 0.0075mol H0.0273M

0.125L 0.150L 0.275LH O H



1L 1mol KOHinitial

The concentration of H+ is equal to moles of H+ per Litre of solution. The moles of H+ is 0.0075 moles.

2 4

2 4


1L 1mol H SOinitial

+Excessmol H 0.0375mol H 0.0300mol 0.0075mol HOH

+ +

3

0.0075mol HH O H 0.0273M

0.125L

0.0075mol H

0.150L 0.275L



1L 1mol KOHinitial

And the total volume of the mixture is 0.125 L of H2SO4 ….

2 4

2 4

0.150mol H SO 2mol Hmol H 0.0375mol0 H

1L 1mol SO5L

H.12initial

+ +

3 0.125L

0.0075mol H 0.0075mol HH O H 0.0273M

0.150L 0.275L



1L 1mol KOHinitial


Plus 0.150 L of KOH.

2 4

2 4


1L 1mol H SOinitial

+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 00.150L .275L


0.200mol KOH 1mol OHmol OH 0.0300mol OH

1L 1mol0.150L

KOHinitial


So the concentration of H+ or H3O+ is 0.0075 moles over 0.275 L.

2 4

2 4


1L 1mol H SOinitial

+ +

3

0.0075mol H0.0273M

0.125L 0.150L

0.0075mol HH O H

0.275L



1L 1mol KOHinitial


Which comes out to 0.0273 molar. We’ll carry one more significant figure than the 2 our final answer is limited to. We’ll round to 2 significant figures at the end.

2 4

2 4


1L 1mol H SOinitial

+ +

3

0.0075mol H 0.0075mol H

0.125L 0.150LH O H 0.0273M

0.275L



1L 1mol KOHinitial


In the last step, we’ll find the pH of the mixture.

2 4

2 4


1L 1mol H SOinitial

+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L

3log H O log 0.0273 1.56pH 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?


1L 1mol KOHinitial


Remember, pH is defined as the negative log of the hydronium ion concentration.

2 4

2 4


1L 1mol H SOinitial

+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L

3 log 0.0273 1.pH log H O 56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?


1L 1mol KOHinitial


Which is the negative log of 0.0273

2 4

2 4


1L 1mol H SOinitial

++

3

0.0075mol H 0.0075mol HH M

0.12H O 0

5L 0.150L 0.275L.0273

3 log 0.02pH log H O 1.5673 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?


1L 1mol KOHinitial


, which comes out to 1.56.

2 4

2 4


1L 1mol H SOinitial

+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L

3log H O log 0.02pH 1.5673 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?


1L 1mol KOHinitial



In a pH, the digits to the right of the decimal are significant. So this answer has 2 significant figures.

2 4

2 4


1L 1mol H SOinitial

+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L

3log H O log 0.02pH 5673 1. 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?


1L 1mol KOHinitial



Now we have answered the original question. The pH of the final mixture is 1.56. This low value means the solution is fairly acidic. This is reasonable because a strong acid is in excess in this case.

2 4

2 4


1L 1mol H SOinitial

+ +

3

0.0075mol H 0.0075mol HH O H 0.0273M

0.125L 0.150L 0.275L

3log H O log 0.02pH 1.5673 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?


1L 1mol KOHinitial


pH of Final Mixture

Documents

Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base