(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