a. 0.1 mol H₂SO₄/1 L = 0.1 M H₂SO₄

H₂SO₄ ® H⁺ + HSO₄^-

so [H⁺] = 0.1 M, [HSO₄^-] = 0.1 M

HSO₄^- Û H⁺ + SO₄^2- Ka. = 0.012

	HSO4^-	H^+	SO4^2-
Intial	0.1	0.1	0
Change	-x	+x	+x
Final	0.1 – x	0.1 + x	x

0.012 = x(0.1+x)/(0.1-x)

0.0012-0.012x = 0.1x + x²

0 = x² + 0.1012x – 0.0012

solve via quadratic

x = 0.011

[H⁺] = 0.11 M

pH = 0.96

b. 0.1 M LiF

LiF ® Li⁺ + F^-

[Li⁺] = [[F^-] = 0.1 M

F^- + H₂O Û HF + OH^- Ka of HF = 7.2 X 10^-4

Kb of F^- = 1.4 X 10^-11

	F^-	OH^-	HF
initial	0.1	0	0
change	-x	+x	+x
final	0.1 – x	x	x

Kb = x²/(0.10 – x) = 1.4 X 10^-11

assume x is small

1.2 X 10^-6 = x = [OH^-]

pOH = 5.92

pH = 8.07

c. 0.1 M C₅H₅N

C₅H₅N + H₂O Û C₅H₅NH⁺ + OH^- Kb = 1.5 X 10^-9

	C5H5N	C5H5N	OH^-
initial	0.1	0	0
change	-x	+x	+x
equilibrium	0.1 - x	x	x

Kb = x²/(0.1-x) = 1.5 X 10^-9

assume x is small

x = [OH^-] = 1.2 X 10^-5

pOH = 4.91

pH = 9.09

d. Ag₃PO₄ is insoluble, so

Ag₃PO₄ Û 3 Ag⁺ + PO₄^3-

Ksp = [Ag⁺]³[PO₄^3-] = 1.3 X 10^-20

if “x” amount of silver phosphate dissolves, then

[Ag⁺] = 3x and [PO₄^3-] = x

1.3 X 10^-20 = 27x⁴

4.7 X 10^-6 = x

1.4 X 10^-5 M = [Ag⁺]

pAg = 4.85

e. 0.1 M KOH

KOH ® K⁺ + OH^-

[OH^-] = 0.1

pOH = 1.0

pH = 13.0

f. 0.1 M Al₂(SO₄)₃

Al₂(SO₄)₃ ® 2 Al³⁺ + 3 SO₄^2-

[Al³⁺] = 0.2 M [SO₄^2-] = 0.3 M

Al(H₂O)₆³⁺ Û H⁺ + Al(H₂O)₅(OH)²⁺ Ka = 1.2 X 10^-5

1.2 X 10^-5 = x²/(0.2 – x)

assume x is small

1.5 X 10^-3 M = x = [H⁺]

pH = 2.81

g. 0.1 M Na₂HPO₄

Na₂HPO₄ ® 2 Na⁺ + HPO₄^-2

[Na⁺] = 0.2 M [HPO₄^-2] = 0.1 M

HPO₄^-2 Û H⁺ + PO₄^3- Ka₃ = 3.6 X 10^-13 pKa = 12.44

HPO₄^-2 + H₂O Û OH^- + H₂PO₄^- Kb = Kw/Ka₂ = 1.0 X 10^-14/6.2 X 10^-8 pKa = 7.21 pKb = 14.00 – 7.21 = 6.79

since it’s amphoteric

pH » (pKa₁ + PKa₂)/2

pH » (6.79 + 12.44)/2 = 9.62

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