Common Ion Effect Problem

CH 117 Spring 2015Worksheet 18

Chapter 15 Review

1. What is the molar solubility of Ag2CO3 in aqueous 0.2 M Na2CO3? The Ksp value for Ag2CO3 is 8.1 x 10-12.

Common ion effect problem

Set up Kap expression and solve for x.

Ksp = 8.1 x 10-12 = (0.2 + x)(2x)2 = 4x2(0.2 + x)  use the 5% rule to ignore the x in parentheses

8.1 x 10-12 = 0.2(4x2) = 0.8x2

x2 = (8.1 x 10-12)/0.8 = 1.0125 x 10-11x = 3.18 x 10-6 M

1. Calculate the pH of a 1 L aqueous solution that is 0.2 M in HCN and 0.3 M in NaCN after 0.011 mol of NaOH is bubbled through the solution. The Ka of HCN is 6.17 x 10-10.

HCN/NaCN solution  BUFFER problem

The NaOH added will react with the acid present in the buffer. NaOH acts as the limiting reagent.

Still have a buffer solution left at the end.

Since we are in 1 L  [HCN] = .189 M and [CN-] = .311 M

To calculate the pH of a buffer we use Henderson-Hasselbach.

pH = -log(6.17 x 10-10) + log(.311/.189) = 9.43

1. What is the pH of a solution that is 0.08 M in aqueous ammonia and 0.040 M in ammonium chloride? The Kb of ammonia is 1.8 x 10-5.

This is a buffer with no additions, so we can use Henderson-Hasselbach directly to calculate pH. NOTICE THE DIFFERENCE BETWEEN THIS PROBLEM AND THE ONE BEFORE IT!

This is a basic buffer, and we are given a Kb. To use the equation, we need a Ka.

Kw = Ka*KbKa = Kw/Kb = (1.0 x 10-14)/(1.8 x 10-5) = 5.56 x 10-10

pH = -log(5.56 x 10-10) + log(.08/.04) = 9.56

1. What is the pH after 37.0 mL of 0.532 M HCl has been titrated with 42.6 mL of 0.203 M KOH?

This is a strong acid + strong base titration, so it becomes simply a limiting reagent problem.

These limiting reagent ICE charts are always worked in moles! (Multiply molarity by the volume).

HCl: .037 L * .532 M = .0197 moles

KOH: .0426 L * .203 M = .00865 moles  limiting reagent

All we have left in solution is strong acid, so we can use that to calculate pH. Don’t forget to divide by the total volume to calculate molarity before you go any further!

[HCl] = .01105 moles/ (.037 + .0426 L) = .139 M

pH = -log(.139 M) = 0.857

1. For each of the following problems: 33.7 mL of 0.434 M C6H5NH2 is titrated with 0.319 M HBr. The Kb of C6H5NH2 is 4.2 x 10-10.

Weak base + strong acid titration

a). What is the pH at the start of the titration before any HBr has been added?

This is essentially a problem from chapter 14. We are asked to find the pH of a solution of weak base. Volume doesn’t matter here because we aren’t mixing solutions together.

Kb = 4.2 x 10-10 = x2/(.434 – x)  use the 5% rule to ignore the x in the denominator

x2 = .434*(4.2 x 10-10)  x = 1.35 x 10-5 this is also the value of [OH-] in our ICE chart, which we can use to calculate pH

pOH = -log(1.35 x 10-5) = 4.87

pH = 14 – 4.87 = 9.13

b). What is the pH at the equivalence point of the titration?

Remember that at the equivalence point of the titration: moles of acid = moles of base.

The first step to any titration problem is figuring out how many moles of each acid and base you have in solution.

Moles of base: .0337 L * .434 M = .0146 moles

Moles of acid: we aren’t given a volume, but since we are at the equivalence point, we should have .0146 moles of acid also.

We will eventually need the volume of acid that we added, but this can easily be figured out by dividing the number of moles by the molarity  V = .0146 moles/.319 M = .0458 L

Set up the first limiting reagent ICE chart.

We are left with only conjugate acid in solution, so we will need another regular ICE chart to complete the calculation. The equation for this chart is the reverse of the one used above, or what happens when we dissolve the conjugate acid in water.

We also need to work this regular ICE chart in molarity (only limiting reagent ICE charts can be done in moles).

[HB+] = .0146 moles/(.0337 L + .0458 L) = .184 M

Since this is an acidic solution, we also need a Ka instead of the Kb given to us.

Ka = Kw/Kb = (1.0 x 10-14)/(4.2 x 10-10) = 2.38 x 10-5

We can now set up a Ka expression to solve for x or [H3O+].

Ka = 2.38 x 10-5 = x2/(.184 – x)  use the 5% rule to ignore the x in the denominator

x2= .184*(2.38 x 10-5)  x = .0021 M = [H3O+]

pH = -log(.0021) = 2.68

c). What is the pH after the addition of 11.3 mL of HBr?

Moles of base: .0337 L * .434 M = .0146 moles

Moles of acid: .0113 L * .319 M = .0036 moles  limiting reagent

We have a buffer solution left over  use Henderson-Hasselbach

First calculate molarities by dividing by the total volume.

[B] = .011 moles/(.0337 +.0113) = .244 M

[HB+] = .0036 moles/(.0337 +.0113) = .08 M

Ka was solved for in part b Ka =2.38 x 10-5

pH = -log(2.38 x 10-5) + log(.244/.08) = 5.11

d). What is the pH at the midpoint of the titration?

Midpoint is another name for the half-equivalence point  point at which pH = pKa

pH = -log(2.38 x 10-5) = 4.62

1. Will a precipitate form when 250 mL of 0.2 M K2SO4 is mixed with 150 mL of 0.1 M Pb(NO3)2? The Ksp value for PbSO4 is 1.8 x 10-8.

We are mixing two salts together in this problem, so we should start by writing an equation that will help us identify the precipitate.

K2SO4 (aq) + Pb(NO3)2 (aq)  2 KNO3 (aq) + PbSO4 (s)

You should be able to write this equation using concepts from last semester. It is a double-displacement reaction, and we can identify the precipitate using our solubility rules.

The precipitate is the only thing we are concerned with!

PbSO4 (s)  Pb2+ (aq) + SO42- (aq)

In order to see whether or not this precipitate will actually form, we need to calculate Q (the reaction quotient) and compare it to the Ksp given.

Q for PbSO4 (s) = [Pb2+][SO42-]

We can calculate these concentrations using the values given to us in the problem. Since we are mixing two solutions together, we must take into account the total volume when calculating concentrations.

[Pb2+] = (.15 L * .1 M)/(.4 L) = .0375 M

[SO42-] = (.25 L * .2 M)/(.4 L) = .125 M

Q = .0375 * .125 = .00469

.00469 (Q) is greater than the Ksp value given, so YES a precipitate will form