What combination of substances will give a buffered solution that has a pH of 5.05?

A) [NH3]= 1.0 /5 = 0.2 M
[NH4 ] = 1.5 / 5 = 0.3 M
NH3 H2O <---> NH4 OH-
1.8 x 10^-5 = (0.3 x)(x) / 0.2-x
x = [OH-] =0.000012 M
pOH = 4.92
pH = 9.08

B) 1.8 x 10^-5 = (0.2 x)(x) / 0.3-x
x = 0.000027 M
pOH = 4.57
pH = 9.43

C)
C5H5N H2O <----> C5H5NH OH-
initial concentration
0.2 .. . . . . . .. . . . . . .. . . 0.3
at equilibrium
0.2-x. . . . . .. . . . . . . . . . 0.3 x .. . .. . x

[C5H5] = 1/ 5 = 0.2 M
[C5H5NH ] = 1.5 / 5 = 0.3 M

1.7 x 10^-9 = (0.3 x)(x) / 0.2-x
x = 1.13 x 10^-9 = [OH-]
pOH =8.95

pH =14 - pOH = 5.05
C is the answer

A)
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) …. Kb = 1.8 × 10⁻⁵
pOH = pKb + log([NH₄⁺]/ [NH₃]) = -log(1.8 × 10⁻⁵) + log(1.5/1.0) = 4.92 ≠ 5.05
pH = pKw - pOH = 14.00 - 4.92 = 9.08

B)
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) …. Kb = 1.8 × 10⁻⁵
pOH = pKb + log([NH₄⁺]/ [NH₃]) = -log(1.8 × 10⁻⁵) + log(1.0/1.5) = 4.57 ≠ 5.05
pH = pKw - pOH = 14.00 - 4.57 = 9.25

C)
C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq) …. Kb = 1.7 × 10⁻⁹
pOH = pKb + log([C₅H₅NH⁺]/ [C₅H₅N]) = -log(1.7 × 10⁻⁹) + log(1.5/1.0) = 8.95
pH = pKw - pOH = 14.00 - 8.95 = 5.05

D)
C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq) …. Kb = 1.7 × 10⁻⁹
pOH = pKb + log([C₅H₅NH⁺]/ [C₅H₅N]) = -log(1.7 × 10⁻⁹) + log(1.0/1.5) = 8.60
pH = pKw - pOH = 14.00 - 8.95 = 5.40 ≠ 5.05

The answer: C) 1.0 mole C5H5N and 1.5 mole C5H5NHCl