(3.c) 1.3 x 10-3 M
x = [H+] = [HAc] This is true because these ions come from HAc, the only source of ions.
| HAc | <---> | H+ | + | Ac- | |
| Initial(I) | 0.10 M | 0 M | 0 M | ||
| Change(C) | -x | +x |
| Equilibrium(E) | (0.10-x)* | x | x |
*Helpful hint: [HAc]i > 100 Ka you can simplify this by assuming that x<< 0.10 M and hence (0.10 - x) ~ 0.10
[HAc]i < 100 Ka you probably can't make the assumption and will need to solve a quadratic equation.
Whenever you make this assumption to simplify the calculation you must go back and check to be sure the assumption is correct (in this case, the x that you calculate is much less than 0.10 M.)
Ka = ([H+][Ac-])/[HAc]
1.8 x 10-5 = {(x)(x)}/0.10
1.8 x 10-6 = x2
x = 1.3 x 10-3 M