The secret to solving these problems is to recognize what you have in solution at the point which the question addresses. Then write an equation and prepare an ICE table. Like the following which I will start but you must finish.
500 mL x 0.178 = 89.0 mmoles NaOH
500 mL x 0.100 = 50.0 mmoles CH3COOH
.....NaOH + CH3COOH ==> CH3COONa + H2O
So you have 39.0 mmoles NaOH after the reaction; therefore, the pH will be determined by the OH^- from NaOH.
(OH^-) = mmoles/mL
mmoles = 39.0 and mL = 1000 mL
(OH^-) = 0.039mols/L = 0.0390 M
The others are done similarly.
I calculated all of them and this is what I got:
1) [H+] = 2.56x10^-13 M
2) [OH-] = .039 M
3) [Na+] .089 M
4) [CH3COO-] = .05
5) [CH3COOH] = 0
All of them were right except number 5 and I don't know what the answer to number 5 is suppose to be.
I would have agreed with all of your answers, including number 5, because whatever it is it must be so small as to be negligible. YOu might try this.
CH3COOH ==> H^+ + CH3COO^-
Ka = (H^+)(CH3COO^-)/(CH3COOH)
If I substitute Ka = 1.8 x 10^-5 (your text may have a different number) and (H^+) from the problem as 2.56E-13 along with (CH3COO^-) = 0.05 from the problem and solve for (CH3COOH), I get something like 7.11E-10 which is, indeed, quite small. Or we could look at it another way, CH3COO^- is a base and hydrolyzes to produce
CH3COO^- + HOH ==> CH3COOH + OH^-
Kb = (Kw/Ka) = (1E-14/1.8E-5) = 5.55E-10 = (CH3COOH)(OH^-)/(CH3COO^-).
If I substitute 0.039 for OH and 0.05 for CH3COO^- and solve for CH3COOH, I get 7.11E-10. Same value. That's a reasonable answer I think and it IS small. Let me know if this doesn't do it.
yes 7.11E-10 was the right answer thank you very much for the help :)
Wait, then why isn't the concentration of [H+] also considered to be zero since the value is so small?