Chapter 15 Applications of Aqueous Equilibria. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 15–2 QUESTION Suppose the weak

Chapter 15

Applications of Applications of Aqueous EquilibriaAqueous Equilibria

Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 15–2

QUESTIONSuppose the weak acid HNO2 (Ka = 4.0 10–4) is added to a solution of NaNO2. If the concentration of acid is 0.10 M and the salt concentration is 0.060 M, what is the [H+]?

1. 2.4 10–4 M2. 6.7 10–4 M3. 2.0 10–4 M4. 4.0 10–5 M


ANSWERChoice 2 represents the approximate [H+]. The Ka expression may be used with the assumption that the dissociation of the acid is very slight. The presence of NO2

– (from the salt) does reduce the amount of dissociation of the acid. The assumption that any NO2

– from the dissociation of the acid can be ignored is used in the calculation so that [H+] = Ka (0.10)/(0.060).

Section 15.1: Solutions of Acids or Bases Containing a Common Ion


QUESTIONThe dissociation constant of propanoic acid (HC3H5O2) is 1.3 10–5. What is the pH of a buffer made with 0.50 M propanoic acid and 0.25 M NaC3H5O2 (the sodium salt of propanoic acid)?

1. 4.592. 5.193. 7.004. I’m not getting any of these answers.


ANSWERChoice 1 is the correct pH of the buffer. There is an assumption being made here about the concentration of the salt to be only from the salt added. This ignores the small amount of propionate [C3H5O2

–] that may be produced from the acid dissociation.

Section 15.2: Buffered Solutions


QUESTIONA certain chemical reaction runs very slow at high pH values and very fast at lower pH values. To study the reaction, a chemist needs to buffer the solution at a basic pH. Determine the pH of a buffer solution that had the following concentrations: 0.42 M NH4Cl and 0.75 M NH3; Kb of NH3 = 1.8 10–5 at 25°C.

1. 9.002. 4.493. 9.514. 11.57


ANSWERChoice 3 provides the correct basic pH of the solution. Using the Kb value of ammonia allows the calculation of [OH–] with the given NH3 and NH4Cl concentrations. Assuming that no significant NH4

+ comes from the NH3 + H2O reaction, the [OH–] could be approximated via [OH–] = Kb (NH3)/(NH4

+); the pH can then be determined from the pOH.



QUESTIONA 0.50 L buffer solution containing 0.42 M NH4Cl and 0.75 M NH3 (Kb of NH3 = 1.8 10–5 ) has a pH of 9.51 at 25°C. The solution receives 0.010 moles of HCl from an outside source. Assuming no significant change in volume of the solution, what is the pH of the solution after the addition of the HCl at 25°C?

1. 9.482. 9.513. 9.344. 8.76


ANSWERChoice 1 correctly incorporates the new concentrations of base and salt after the addition of H+. The concentrations of both base and its salt can be converted to amounts by using V M = moles. Then the moles of NH3 are reduced by 0.010 moles of H+

from the HCl. This means that the NH4+ amount is increased by

the same 0.010 moles of H+. The new amounts are then divided by the solution volume (0.50 L) to determine the new concentrations. These are then used to solve for the [OH–], which is converted to pOH and finally to pH.



QUESTIONAlthough buffers resist change in pH they are still affected by the addition of a new base or acid. A 0.50 L buffer solution containing 0.42 M NH4Cl and 0.75 M NH3;(Kb of NH3 = 1.8 10–5 ) has a pH of 9.51 at 25°C. What is the new pH of the buffer after it is adjusted by the addition of 0.010 moles of NaOH. Assume no change in the solution volume.

1. 8.972. 9.513. 12.004. 9.54


ANSWERChoice 4 provides the newly adjusted pH. The addition of 0.010 moles of new OH– causes the NH4

+ concentration to decrease and the NH3 concentration to increase. The molarity of both is first changed to moles (V M = moles;) then the 0.010 moles of OH– is added and subtracted appropriately. The new number of moles of each are divided by the solution volume to obtain molarity. The system can now be treated with the Kb expression using the new molarities of the NH4

+ and NH3 to solve for [OH–], which can then be used to obtain pOH and pH.



QUESTIONTwo separate buffer solutions are prepared using propanoic acid and calcium propanoate (also used in some food preservative applications). If buffer “A” was 0.10 M in both acid and salt while buffer “B” was 0.20 M in both acid and salt, which of the following would be true?

1. Both solutions would have the same pH and would havethe same buffer capacity.

2. Solution “B” would would have a lower pH and both would have the same buffer capacity.3. Solution “B” would have a lower pH and a larger buffer

capacity.4. Both solutions would have the same pH; “B” would have

a larger buffer capacity.


ANSWERChoice 4 offers the accurate comparison of the pH and buffer capacity for the two solutions. As the salt and acid have the same concentration, the pH = pKa relationship is the same. However, because solution “B” has a higher molarity for both acid and salt any additional acid or base additions will have to be much larger before the ratio changes significantly. Therefore, solution “B” has a higher buffering capacity.

Section 15.3: Buffering Capacity


QUESTIONTo culture a certain bacteria a microbiologist would like to buffer the media at a pH of 3.75. To maximize the efficiency of the system a 1:1 ratio of acid to salt will be used. Which of the following acids would make the best choice for the buffer?

1. Acetic acid; Ka = 1.8 10–5

2. Propanoic acid; Ka = 1.3 10–5

3. Formic acid; Ka = 1.8 10–4

4. Nitrous acid; Ka = 4.0 10–4


ANSWERChoice 3 represents the best selection for this buffer. If the acid:salt ratio is to be 1:1, the pH will equal pKa. At a pH of 3.75 the closest pKa would be from formic acid: 1.3 10–4.

Section 15.3: Buffering Capacity


QUESTIONIf 25.0 mL of 0.250 M HCl were titrated with 0.500 M NaOH, what are the initial pH, pH at the equivalence point and volume of the total solution at the equivalence point?

1. 1.000; 7.00; 25.0 mL 2. 0.602; 7.00; 50.0 mL3. 0.602; 7.00; 37.5 mL4. 1.000; 7.00; 12.5 mL


ANSWERChoice 3 provides the correct response for all three questions. The pH of a strong acid is taken directly from the molarity. (–log 0.250). The pH of the equivalence point for a strong acid and strong base titration is 7.00. The moles of HCl initially is 25.0 mL 0.250 M, or 6.25 millimoles. Therefore, the base must provide 6.25 millimoles for neutralization. The volume of 0.500 M NaOH that would provide this could be calculated from V 0.500 M = 6.25; V = 6.25/0.500 = 12.5 mL.

Section 15.4: Titrations and pH Curves


QUESTIONWhat is the pH of the equivalence point of the titration of 30.0 mL of 0.100 M benzoic acid, C6H5CO2H (Ka = 6.4 10–5), with 0.100 M NaOH at 25°C?

1. 8.602. 5.553. 8.454. I must be missing a step, I don’t get any of these.


ANSWERChoice 3 correctly accounts for the formation of the moles of the salt of benzoic acid during the titration (volume acid molarity of the acid), molarity of the salt (moles / total volume), determination of the Kb of the conjugate salt (Kw/Ka), and finally the pH (14 – pOH).



QUESTION50.0 mL of an acid solution (HA) contains 0.0200 mole of the acid. The solution is titrated with 0.200 M NaOH. Half way to the equivalence point the pH is 5.25. What is the total volume of the solution at that point and what is the Ka of the acid?

1. 0.100 L; Ka = 5.6 10–6

2. 0.150 L; Ka = 5.6 10–6

3. 0.100 L; Ka = 1.8 10–9

4. 0.150 L; Ka = 1.8 10–9


ANSWERChoice 1 provides both the correct volume of the solution and correct Ka value for the acid. 50.0 mL of acid containing 0.0200 mole would require 0.100 L of 0.200 M NaOH to neutralize, so half way to that value would be 50.0 mL. This is then added to the original 50.0 mL of acid to obtain the solution volume of 0.100 L half way to the equivalence point. At this point the concentration of HA is approximately that of A–, so Ka = [H+] or pKa = pH.



QUESTIONSuppose you have 50.0 mL each of 0.10 M solutions of propanoic acid (Ka = 1.3 10–5); acetic acid(Ka = 1.8 10–5); HCl (all monoprotic acids). Which would require the LEAST volume of 0.10 M NaOH to neutralize?

1. Propanoic acid2. Acetic acid3. HCl4. I think they all require the same.


ANSWERChoice 4 is correct. As all the acids have the same volume and molarity, they have the same number of moles. NaOH is a strong base and will react with all the moles of each acid. Therefore, each would require the same moles of NaOH for neutralization.



QUESTIONMost acid-base indicators are weak acids. In a titration of 0.50 M acetic acid (at 25°C, Ka = 1.8 10–5) with KOH, which indicator would best indicate the pH at the equivalence point? The approximate Ka for each choice is provided.

1. Bromophenol blue; Ka ~ 1 10–4

2. Methyl red; Ka ~ 1 10–5

3. Bromothymol blue; Ka ~ 1 10–7

4. Thymolphthalein; Ka ~ 1 10–10


ANSWERChoice 4 provides the best choice although there may also be better choices available than these four. The equivalence point pH should be as close as possible to the pKa of the indicator. As acetic acid is a fairly weak acid and NaOH is a strong base, the pH at the equivalence point will be above 7. The only choice above 7 in the list was thymolphthalein. Without a more detailed calculation, this would be the best choice.

Section 15.5: Acid–Base Indicators


QUESTIONThe acid-base indicator bromcresol purple has an interesting yellow-to-purple color change. If the approximate Ka of this indicator is 1.0 10–6, what would be the ratio of purple [A–] to yellow [HA] at a pH of 4.0?

1. 100:12. 1:1003. 1:14. This choice indicates that I don’t know.


ANSWERChoice 2 shows the [A–]/[HA] ratio at pH 4.0 for bromcresol purple. The pH can be converted to [H+] and divided into the Ka value to reveal the [A–]/[HA] ratio at pH 4.0.Ka/[H+] = [A–]/[HA].

Section 15.5: Acid–Base Indicators


QUESTION

Lead (II) iodide is used in some camera batteries. PbI2 has a Ksp of 1.4 10–8. What is the molar solubility of this compound?

1. 1.9 10–3 M2. 2.4 10–3 M3. 1.5 10–3 M4. 8.4 10–5 M


ANSWERChoice 3 correctly uses the stoichiometry and equilibrium expression to determine the molar solubility of PbI2. Since the dissociation of PbI2 shows Pb2+(aq) + 2I–(aq), the Ksp expression would appear as: Ksp = [X][2X]2. Solving for X shows the relationship for PbI2 dissociating in a 1:1 ratio between PbI2 and Pb2+.

Section 15.6: Solubility Equilibria and the Solubility Product


QUESTIONCalcium phosphate (Ca3(PO4)2) is only slightly soluble. In fact, it is a common ingredient in phosphate rock and is a major source of phosphate fertilizer. If the molar solubility is 2 10–7, what is the value of the Ksp?

1. 3 10–34

2. 2 10–10

3. 3 10–32

4. I don’t get any of those values.


ANSWERChoice 3 correctly incorporates the stoichiometry and equilibrium expression for calcium phosphate. Ksp = [Ca2+]3[PO4

3–]2. Since the molar solubility represents the dissociation of Ca3(PO4)2, the ratio of that to the ions would be 1:3 and 1:2 respectively. So, the expression becomes: Ksp = [3 (2 10–7)]3[2 (2 10–7)]2.



QUESTION

The Ksp of Ag2CrO4 is 9.0 10–12. What would be the solubility of this compound in a solution that was already 0.10 M in potassium chromate?

1. 9.0 10–13 M2. 9.5 10–6 M3. 3.0 10–6 M4. 4.7 10–6 M


ANSWERChoice 4 provides the correct solubility of the compound in a solution with a common ion. Ksp = [Ag+]2[CrO4

2–]; The [Ag+] = 2 the molar solubility. Proper substitution of known values produces:

9.0 10–12 = [Ag+]2[~ .10]



QUESTIONWill a precipitate of BaSO4 form when 10.0 mL of 0.0010 M barium nitrate are mixed with 20.0 mL of 0.000020 M of sodium sulfate? The Ksp of barium sulfate is 1.5 10–9. Prove your answer by reporting the calculated value of Q.

1. Yes; Q = 2.0 10–8

2. Yes; Q = 4.4 10–9

3. No; Q = 7.0 10–10

4. No; Q = 3.0 10–17


ANSWER

Choice 2 provides the correct prediction. Once the molarity of each ion (Ba2+ and SO4

2–) is determined they are multiplied to obtain Q. Since Q is larger than Ksp a precipitate will form.

Section 15.7: Precipitation and Qualitative Analysis

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Chapter 15 Applications of Aqueous Equilibria. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 15–2 QUESTION Suppose the weak