Question

A solution consists of 3.712 M C5H5N and 2.468 M C5H5NHBr. (Kb = 1.40 x 10^-9 for C5H5N)

a) Calculate pH. Using i.c.e chart.

b) Calculate the pH after 1500 mL of 3.554 M HCl are added to 535.0 mL of the buffered solution above, in part a.

Answer 1

a) The reaction of base with water is as below.

C₅H₅N (aq) + H₂O (l) ------> C₅H₅NH⁺ (aq) + OH^- (aq)

Initial                3.712                                        2.468                 0

Change                -x                                             +x                  +x

Equilibrium    (3.712 – x)                             (2.468 + x)             x

Write down the expression for K_b.

K_b = [C₅H₅HN⁺][OH^-]/[C₅H₅NH]

===> 1.40*10^-9 = (2.468 + x).(x)/(3.712 – x)

We need to make an assumption here. K_b is extremely small (of the order of 10^-9); hence we can assume x << 2.468, 3.712 and therefore, we can assume (3.712 – x) ≈ 3.712 and (2.468 + x) ≈ 2.468. Therefore,

1.40*10^-9 = 2.468*x/3.712

===> 2.468*x = 3.712*1.40*10^-9

===> x = 5.5830*10^-8

Therefore, [OH^-] = 5.5830*10^-8 M and pOH = -log [OH^-] = -log (5.5830*10^-8) = 7.25; therefore, pH = 14.00 – pOH = 14.00 – 7.25 = 6.75 (ans).

b) pK_b = -log K_b = -log (1.40*10^-9) = 8.8539

Moles of C₅H₅N in the buffer = (volume of buffer)*(concentration of C₅H₅N) = (535.0 mL)*(3.712 mol/L) (1 M = 1 mo/L) = 1985.92 mmole.

Moles of C₅H₅NH⁺ = (volume of buffer)*(concentration of C₅H₅HN⁺) = (535.0 mL)*(2.468 mol/L) = 1320.38 mmole.

Moles of HCl (H⁺) added = (15.00 mL)*(3.554 mol/L) = 53.31 mmole (I believe its 15.00 mL and not 1500 mL)

HCl will neutralize C₅H₅N as per the equation below.

C₅H₅N (aq) + H⁺ (aq) -------> C₅H₅NH⁺ (aq)

The solution contains both C₅H₅N and C₅H₅NH⁺.

Moles of C₅H₅N = (initial moles of C₅H₅N) – (moles of HCl added) = (1985.92 – 53.31) mmole = 1932.61 mmole.

Moles of C₅H₅NH⁺ = (initial moles of C₅H₅NH⁺) + (moles of HCl added) = (1320.38 + 53.31) mmole = 1373.69 mmole.

Total volume of solution = (1500 + 535.0) mL = 2035.0 mL.

[C₅H₅N] = (1932.61 mmole)/(2035.0 mL);[C₅H₅NH⁺] = (1373.69 mmole)/(20.35.0 mL)

Use the Henderson-Hasslebach equation for bases.

pOH = pK_b + log [C₅H₅NH⁺]/[C₅H₅N] = 8.8539 + [(1373.69/2035.0 mmol/mL)/(1932.61/2035.0 mmol/mL)] = 8.8539 + log (1373.69/1932.61) = 8.8539 + log (0.7108) = 8.8539 + (-0.1482) = 8.7057 ≈ 8.70

pH = 14.00 – pOH = 14.00 – 8.70 = 5.30 (ans).