Homework SourseWhat would the pH of this solution be if enough NaCl were ad
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+
Solution
Ka = yH+[H+] * yOAc-[OAc-]/[HOAc]
1.8*10^-5 = yH+[H+] * yOAc-[OAc-]/[HOAc]
get activity coffieicents
Recall that:
-log() = 0.51*(Zi^2)*sqrt(I.S.) / ( 1 + ( * sqrt(I.S)/305))
Where
i = activity coefficient for species “i”
i = theoretical diameter in pm (10^-12 m)
Zi = Charge of ion
I.S. = ionic Strength (usually used as as well)
If we wanted only
= 10^-(0.51*(Zi^2)*sqrt(I.S.) / ( 1 + ( * sqrt(I.S)/305)))
for H+ , Zi = +1, IS = 0.1, = 900
YH+ = 10^-(0.51*(1^2)*sqrt(0.1 ) / ( 1 + (900* sqrt(0.1 )/305))) = 0.8252
for OAc- , Zi = -1 , IS = 0.1, = 450
YOAc- = 10^-(0.51*(1^2)*sqrt(0.1 ) / ( 1 + (450* sqrt(0.1 )/305))) = 0.7763
1.8*10^-5 = yH+[H+] * yOAc-[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(yH+[H+] )
pH = -log(0.8252*0.00166)
pH = 2.8633
