Question

What is the pH of a 1.3 x 10-7 M solution of HCl? Please note that you cannot simply add the contribution of H+ ions from the autoproteolysis of water and the contribution from the dissociation of HCl. You will need to use the quadratic equation to solve this problem and report the answer to 4 significant figures.

Answer 1

What is the pH of a 1.3 x 10-7 M solution of HCl?

Note that HCl is a stong acid

HCl will dissociate 100% to form H+ and Cl-

and H2O <->H+ + OH- equilibirum must be considered since:

H+ concentration is too small

[H₃O⁺] = [OH⁻] + [Cl⁻]

note that [C-] = 1.3*10^-7

[H₃O⁺] = [OH⁻] +1.3*10^-7

Recall Equilibrium of water:

Kw = 10^-14

Kw = [H₃O⁺][OH⁻]

substitute known values

Kw = ([OH⁻] + 1.3*10^-7) * [OH⁻]

10^-14 = [OH⁻]^2 + (1.3*10^-7)*[OH⁻]

which is similar to a quadratic eqution

let x = [OH-] so

x^2 + (1.3*10^-7)x - 10^-14 = 0

solve for x (OH-)

x = 5.426*10^-8

so

[H₃O⁺] = [OH⁻] + 1.3*10^-7 = 5.426*10^-8 + 1.3*10^-7 = 1.842*10^-7

so

pH = -log(H3O+) = -log(1.842*10^-7) = 6.7347

ph = 6.7347 --> 4 fig --> 6.735