Question

What would the pH of this solution be if enough NaCl were added to make a solution with an ionic strength of 0.10 M?

HOAc + H2O -> OAc- + H3O+

Answer 1

Ka = y_H+[H+] * y_OAc-[OAc-]/[HOAc]

1.8*10^-5 = y_H+[H+] * y_OAc-[OAc-]/[HOAc]

get activity coffieicents

Recall that:

-log(γ) = 0.51*(Zi^2)*sqrt(I.S.) / ( 1 + (α * sqrt(I.S)/305))

Where

γi = activity coefficient for species “i”

αi = theoretical diameter in pm (10^-12 m)

Zi = Charge of ion

I.S. = ionic Strength (usually used as μ as well)

If we wanted only γ

γ = 10^-(0.51*(Zi^2)*sqrt(I.S.) / ( 1 + (α * sqrt(I.S)/305)))

for H+ , Zi = +1, IS = 0.1, α = 900

Y_H+ = 10^-(0.51*(1^2)*sqrt(0.1 ) / ( 1 + (900* sqrt(0.1 )/305))) = 0.8252

for OAc- , Zi = -1 , IS = 0.1, α = 450

Y_OAc- = 10^-(0.51*(1^2)*sqrt(0.1 ) / ( 1 + (450* sqrt(0.1 )/305))) = 0.7763

1.8*10^-5 = y_H+[H+] * y_OAc-[OAc-]/[HOAc]

assume [HOAc] = 0.1 M

1.8*10^-5 = 0.8252*x * 0.7763 *(x) /(0.1-x)

x^2/(0.1-x) = (1.8*10^-5) /(0.8252*0.7763 ) = 0.00002809

x^2 = -0.00002809x +0.1*0.00002809

x^2 + 0.00002809x - 0.000002809= 0

x = 0.00166

pH = -log(y_H+[H+] )

pH = -log(0.8252*0.00166)

pH = 2.8633