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Review Unit #13

Chapter #15

1. Devise a scheme to separate a mixture of Ag+ , Ba 2+ and Zn 2+.

• To separate out the silver add hydrochloric acid until all the silver precipitates out as silver chloride

• Now you have a mixture of barium and zinc ions• Add sulfuric acid to the mixture to precipitate out the

barium as barium sulfate, leaving zinc ions alone in solution

Ag+ and Ba2+ and Zn2+

Add HCl

AgCl

Ba 2+ and Zn 2+

Add H2SO4

BaSO4 Zn2+ in solution

2. A student is asked to prepare a buffer solution at pH = 8.60, using one of the following weak

acids: HA (Ka = 2.7 x 10 – 3 ), HB (Ka = 4.4 x 10 – 6 ),

HC (Ka = 2.6 x 10 – 9 ). Which acid should she choose? Why?

• Acid HC. Acid HC has a pKa almost equal to the pH you are seeking.

• When the pH and pKa are almost equal then equal amounts of the acid and the conjugate base give you the right buffer.

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3. A buffer is prepared by mixing 525 mL of 0.50 M formic acid, HCHO2, and 475 mL of 0.50 M sodium formate, NaCHO2. Ka =

1.8 x 10 – 4 .a. Calculate the pH of this buffer solution.

pKa = 3.74 pH = pKa + Log [CHO2-]

[HCHO2][HCHO2] = (.525 L)(0.50 M) = 0.263 M

(1.00 L)

[CHO2-] = (.475 L)(0.50 M)= 0.238 M

(1.00 L)

pH = 3.74 + Log (.238)/(.263) = 3.70

B. What is the pH of 85 mL of the buffer to which 8.5 mL of 0.15 M hydrochloric acid

has been added?CHO2

- + H3O+ HCHO2 + H2O[CHO2

-] = (.085 L)(0.238 M) = 0.0202 mol[HCHO2] = (.085 L)(0.263 M) = 0.0224 mol[H3O+] = (0.0085 L)(0.15M) = 0.00128 mol

[CHO2-] = (0.0202 - .00128) = 0.202 M

(0.0935 L)

[HCHO2] = (0.0224 + .00128) = 0.253 M

(0.0935 L)

pH = 3.74 + Log (.202)/(.253) = 3.64

4. A 1.24 gram sample of benzoic acid, HC7H5O2, is dissolved in water to give 50.0 mL of solution. Ka for benzoic acid is 6.5 x 10 – 5. This solution is titrated with 0.180 M sodium hydroxide, NaOH, solution. Calculate the pH of the resulting solution:

a. after 15.0 mL of NaOH has been added.1.24 g 1 mol = .0102 mol / 0.050 L = .203 M

122 g(0.050L)(.203 M) = .0102 mol Acid(0.0150L)(.180 M) = .00270 mol Base

XS acid = 0.00750 mol Acid / 0.065 L = 0.115 MConjugate Base = .00270 mol/.065 L = 0.0415 M

Ka =[H+][base]

[acid-base]

pH = 4.18 + log [.0415]/[.115] = 3.74pKa =4.18

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4. A 1.24 gram sample of benzoic acid, HC7H5O2, is dissolved in water to give 50.0 mL of solution. Ka for benzoic acid is 6.5 x 10 – 5. This solution is titrated with 0.180 M sodium

hydroxide, NaOH, solution. Calculate the pH of the resulting solution:

b. at the halfway point of the titration.pH = pKapH = 4.18

4. A 1.24 gram sample of benzoic acid, HC7H5O2, is dissolved in water to give 50.0 mL of solution.

Ka for benzoic acid is 6.5 x 10 – 5. This solution is titrated with 0.180 M sodium hydroxide, NaOH,

solution. Calculate the pH of the resulting solution:

c. at the equivalence point.

Kb = Kw/Ka = 1.54 x 10-10

mL base = 50 mL A .203 Mol A 1 mol B 1L B1 L A 1 mol A .180 mol B

mL base = 56.4 mL[base] = (.0564 L)(.180 M) = 0.0954 M

(0.1064L)

Kb = x2/0.0954 M

X = [OH-] =

3.83 x 10-6 M

pH 8.58

A 1.24 gram sample of benzoic acid, HC7H5O2, is dissolved in water to give 50.0 mL of solution. Ka for benzoic acid is 6.5 x 10 – 5. This solution is titrated with 0.180 M sodium

hydroxide, NaOH, solution. Calculate the pH of the resulting solution:

d. After 60.00 mL of NaOH is added

Acid present = 0.0102 mol

Base preset = 0.0108 mol

XS [OH-] = 0.0006 mol/ .110 L = .00545 M

pOH = 2.26 pH = 11.74