Question

Be sure to answer all parts. A solution is made by mixing exactly 500 mL of 0.146 M NaOH with exactly 500 mL of 0.100 M CH3COOH.

Calculate the equilibrium concentration of the species below.

Ka of CH3COOH is 1.8 × 10−5

[H+] M Enter your answer in scientific notation.

[OH−] M

[CH3COOH] M Enter your answer in scientific notation.

[Na+] M

[CH3COO−] M

Answer 1

concentration of NaOH = 500 x 0.146 / (500 + 500) = 0.073 M

concentration of CH3COOH = 500 x 0.1 / 500 + 500 = 0.050 M

[Na+] = 0.0730 M

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

0.050                0.073                                    0                    0

   0                     0.023                               0.050

[NaOH] = 0.0230 M

[OH-] = 0.0230 M

[H+] = 10^-14 / 0.0230

[H+] = 4.35 x 10^-13 M

[CH3COO-] = 0.050 M

CH3COO-   +   H2O   ---------------> CH3COOH +   OH-

Kb = [CH3COOH][OH-] / [CH3COO-]

5.56 x 10^-10 = [CH3COOH] x 0.0230 / 0.050

[CH3COOH] = 1.21 x 10^-9 M