Hi,

Question

I am having trouble with two questions.

Find the pH of the following aqueous solutions prepared by adding:

a) 20mL 0.12M HCl to 10mL 0.16M CH3COOH + 20mL 0.20 KOH.

b) 10mL 5.0x10^-7 M HCl to 90mL H2O

For a), I thought it was a buffer solution so what I tried to do was find the initial moles and add or subtract the moles of HCl (as it reacts with the conjugate base to create more conjugate acid). But then I realised that CH3COOH and KOH were not conjugate pairs. So what do I do?

For b), I tried using M1V2 = M2V2 but the sheet said that "the trivial answer to this is not correct". So how do I do this ?

Thanks for your time.

Answer 1

For a, neutralize HCl with KOH (and there should be KOH left over) and use the excess KOH to neutralize CH3COOH. I think the mmoles left over from the first neutralization will be just the amount to neutralize the CH3COOH. So what you will have in the solutin is KCl (which doesn't affect the pH), H2O from both neutralizations (which is neutral), and CH3COOK (potassium acetate). The acetate ion )a base) is hydrolyzed like so
CH3COO^- + HOH ==> CH3COOH + OH^-

Kb = (K2/Ka) = (CH3COOH)(OH^-)/(CH3COOK).
Set up an ICE chart and substitute into the Kb equation and solve for OH^-, convert to pOH, then to pH. For CH3COOK, that will be the moles formed from the appropriate neutralization divided by the volume of ALL of the solutions.

For b, the trivial answer is not correct because the concn of the HCl given is so SMALL. Remember that the concn of H and OH in water is 1 x 10^-7 M. We usually neglect that when working with solutions BUT when the added soln is about the same, it is no longer negligible. So the total H^+ will be the sum of H^+ from HCl + H^+ from the H2O (which by the way will not be 1 x 10^-7 M).

Answer 2

Thanks,

But for the second one.
What will [H+] be if its not 1x10^-7 ?