A 20.0 mL sample of 0.20 M sodium acetate is titrated with 0.11 M HCl)aq). What is pH after the addition of50.0 mL HCl(aq)? Kbof CH3CO2^- = 5.6x10^-10)

Question

So far I have the

Initial amount of sodium acetate (0.02)(0.20) = .004 mol

Amount of HCl added (0.05)(0.11) = 0.0055 mol

Amount of sodium aceatate after reaction(0.04) - (0.0055) = -.0015

So there is less than nothing left. That can't be right

Answer 1

You are ok so far except that you wrote (0.04)-(0.0055) = -.0015 when you should have written
(0.004)NaAc - (0.0055)HCl = but I think that was just a typo. I have typed in the NaAc and HCl part to clarify things. So what has happened is that you have no NaAc left (technically no Ac^- left except what little may be present from the ionization of a weak acid like acetic acid--which is decreased even further by the addition of the strong acid, HCl). So in effect you have 0.0015 moles HCl in 70 mL water (0.070 L) which means the molarity of HCl is ?? and pH is ??.

Answer 2

Molarity = mol/L

0.0015/.070 = 0.0214

-log 2.14x10^-2 = 1.67