Question

3. The pH of a 0.345 M weak base solution is 9.39. 50.0 mL of the weak base are titrated with 0.425 M HCI. a. Calculate the Kb of the weak base. b. Calculate the Ka of the weak base. c. d. Consult a Chemistry text and determine the identity of the weak base. Calculate the volume of HCl required to reach the equivalence point.

Answer 1

QUESTION 3

a) First we calculate the OH^- concentration using the following equation.

$14-pH=-log[OH^-]$

$14-9.39=-log[OH^-]$

$[OH^-]=2.45*10^{-5}M$

Consider the following equilibrium between the base B and water.

$B+H_2O\rightleftharpoons BH^++OH^-$

Note that the concentrations of BH⁺ and OH^- are equal. The equation for Kb can be written as,

$K_b=\frac{[BH^+][OH^-]}{[B]}=\frac{[OH^-]^{2}}{[B]}$

Substituting the concentrations,

$K_b=\frac{(2.45*10^{-5}M)^{2}}{0.345M-2.45*10^{-5}M}$

We assume that 0.345M-2.45*10^-5M is approximately equal to 0.345M.

$K_b=\frac{(2.45*10^{-5}M)^{2}}{0.345M}=1.74*10^{-9}M$

b) Consider the above equilibrium,

$B+H_2O\rightleftharpoons BH^++OH^-$

BH⁺ can dissociate to give,

$BH^+\rightleftharpoons B+H^+$

Therefore, the Ka equation can be given as,

$K_a=\frac{[B][H^+]}{[BH^+]}$

For a conjugate acid-base pair, we can write,

$K_a*K_b=K_w$

$K_a=\frac{K_w}{K_b}$

$K_a=\frac{10^{-14}}{1.74*10^{-9}}=5.75*10^{-6}$

c) Online sources state that the Kb value of pyridine is 1.7*10^-9. Thus the identity is most likely to be pyridine.

d) The following equation can be used to determine the volume required to reach the equivalence point.

$C_{1}V_{1}=C_{2}V_{2}$

$C_{HCl}V_{HCl}=C_{base}V_{base}$

$0.425M*V_{HCl}=0.345M*50.0ml$

$V_{HCl}=40.588ml=40.6ml$