Question

What is the equilibrium pH of an initially 4.4 M solution of hypochlorous acid, HOCl, at 25°C (Ka = )? What is Ka for a weak monoprotic acid if a 0.020 M solution of the acid has a pH of 3.28 at 25°C? 15.0 mL of 0.50 M HCl is added to a 100.0-mL sample of 0.484 M HNO2 (Ka for HNO2 = 4.0 × 10–4). What is the equilibrium concentration of NO2– ions?

Answer 1

What is the equilibrium pH of an initially 4.4 M solution of hypochlorous acid, HOCl, at 25°C (Ka = )?

HClO = 4*10^-8

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 = 4.4 M; then

x^2 + (4*10^-8)x - 4.4*(4*10^-8) = 0

solve for x

x =4.195*10^-4

substitute

[H+] = 0 + 4.195*10^-4= 4.195*10^-4 M

pH = -log(H+) = -log(4.195*10^-4) = 3.377

pH = 3.377