Question

A 200.0 mL sample of 0.20 MHF is titrated with 0.10 MNaOH. Determine the pH of the solution before the addition of any NaOH. The Ka of HF is 3.5×10−4.

Answer 1

First, assume the acid:

HF

to be HA, for simplicity, so it will ionize as follows:

HA <-> H+ + A-

where, H+ is the proton and A- the conjugate base, HA is molecular acid

Ka = [H+][A-]/[HA]; by definition

initially

[H+] = 0

[A-] = 0

[HA] = M;

the change

initially

[H+] = + x

[A-] = + x

[HA] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = M - x

substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = 0.2 M; then

x^2 + (3.5*10^4)x - 0.2 *(3.5*10^4) = 0

solve for x

x =0.008193

substitute

[H+] = 0 + 0.008193= 0.008193M

pH = -log(H+) = -log(0.008193) = 2.086