(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 K_{a} you can simplify this by assuming that x<< 0.10 M and hence (0.10 - x) ~ 0.10
[HAc]_{i} < 100 K_{a} 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} = x^{2}
x = 1.3 x 10^{-3} M