(5.b) 0.10 M
HCl dissociates completely so that there will be a [H^{+}] = 0.10 M due to the HCl.
This is not completely impacted by the other acid that is present.
HCl(aq) | ----> | H^{+}(aq) | + | Cl^{-}(aq) | |
Initial(I) | 0.10 M | 0 M | 0 M | ||
Change(C) | -0.10 M | +0.10 M | +0.10 M |
Equilibrium(E) | 0 M | 0.10 M | 0.10 M |
There is also [H^{+}] from HAc . The total H^{+} in solution comes from the weak acid (HAc) which is affected by the presence of the common ion H^{+} from the HCl.
HAc(aq) | <---> | H^{+}(aq) | + | Ac^{-}(aq) | |
Initial(I) | 0.10 M | 0.10M^{*} | 0 M | ||
Change(C) | -x | +x | +x |
Equilibrium(E) | (0.10-x)^{*} | (0.10+x)^{*} | x |
*Helpful hint: [HAc]_{i} > 100 Ka you can simplify this by assuming that x<< 0.10 M and hence (0.10 - x) ~ .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} = {(0.10)(x)}/0.10
x = 1.8 x 10^{-5} M = [Ac^{-}] and the contribution of H^{+} from the acetic acid. This [Ac^{-}] is even less than the [H^{+}] from HAc by itself (refer to problem 3). The dissociation of Hac is suppressed by the common ion, H^{+} The total concentration of H^{+} is the sum of the H^{+} from each of the acids:
From the HCl: [H^{+}] = 0.10 and from the HAc the [H^{+}] contribution is 1.8 x 10^{-5} M and the sum of (.10 + 1.8 x 10^{-5}) M = 0.10 M