Question

Calculate the pH of an aqueous solution that's initally 5.0 x 10^-8 M in HCl and 0.10 M in NaCl.

Answer 1

5.0 x 10^-8 M in HCl and 0.10 M in NaCl.

Note that HCl and NaCl are strong electrolytes, will not hydrolyse in water, will remain ionic

so

[H+] = [HCl] = 5*10^-8

then

we can't simply apply

pH = -log(H+) since it will give a pH > 7, which is basic, which is NOT the case

so

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-] = 5*10^-8

[H₃O⁺] = [OH⁻] + 5*10^-8

Recall Equilibrium of water:

Kw = 10^-14

Kw = [H₃O⁺][OH⁻]

substitute known values

Kw = ([OH⁻] + 5*10^-8 ) * [OH⁻]

10^-14 = [OH⁻]^2 + ( 5*10^-8 )*[OH⁻]

which is similar to a quadratic eqution

let x = [OH-] so

x^2 + (5*10^-8)x - 10^-14 = 0

solve for x (OH-)

x = 7.81*10^-8

so

[H₃O⁺] = [OH⁻] + 8*10^-8 = 7.81*10^-8+ 5*10^-8 = 1.281*10^-7

so

pH = -log(H3O+) = -log(1.281*10^-7) = 6.892

ph = 6.892