Question

A solution is made by mixing exactly 500.0 mL of 0.167 M NaOH with exactly 500.0 mL of 0.100 M CH₃COOH. Calculate the equilibrium concentrations of H₃O⁺, CH₃COOH, CH₃COO^-, OH^-, and Na⁺.

[H₃O⁺] = M

[CH₃COOH] =    M

[CH₃COO^-] =    M

[OH^-] = M

[Na⁺] = M

Answer 1

milli moles of NaOH = 500 x 0.167 = 83.5

milli moles of CH3COOH = 500 x 0.100 = 50

CH3COOH + NaOH   --------------------> CH3COONa + H2O

        50             83.5                                   0 0

0 33.5                                        50                 50

equilibrium concentrations:

[CH₃COOH] = 0 M

[CH₃COO^-] = milli moles of CH3COONa / total volume   

                   = 50 / (500 + 500)

                   = 0.05 M

[OH^-] = milli moles of NaOH / total volume   

         = 33.5 / 1000

         = 0.0335 M

[Na⁺] = milli moles of CH3COONa + milli moles of NaOH / total volume

         = 50 + 33.5 / 1000

         = 0.0835 M

[H3O+] = Kw / [OH-]

             = 1.0 x 10^-14 / 0.0335

             = 2.98 x 10^-13 M