Question

A solution is made by mixing exactly 500 mL of 0.167 M NaOH with exactly 500 mL of 0.100 M CH3COOH. Calculate the equilibrium concentrations of H+, CH3COOH, CH3COO-, OH-, and Na+.

Answer 1

mmols = millimols = mL x M.
mmols CH3COOH = 500 x 0.1M = 50
mmols NaOH = 500 x 0.167 = 83.5
Total volume = 500 + 500 = 1000 mL

....CH3COOH + Na^+ + OH^- --> CH3COO^- + Na^+ + H2O
I......50.....83.5........0........0
C.....-50.....-50........+50......+50
E......0......33.5.......50........50

You need to recognize what you have. You have an excess of NaOH, no CH3COOH, you've formed 50 of the salt.

(CH3COOH) = 0
(CH3COO^-) = 50mmols/1000 mL = ?
(Na^+) = never changed. You had 83.5 mmols initially and you have 83.5 mmols at the end. So that's 83.5mmols/1000 mL = ?
(OH^-) comes from the excess NaOH which is 83.5 mmols/1000 mL = ?
Then (H^+)(OH^-) = Kw = 10^-14. You know OH and Kw, solve for H^+.
Check my thinking.