(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

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