Question

Calculate the pH of a 0.800 M NaCH3CO2 solution. Ka for acetic acid, CH3CO2H, is 1.8 × 10-5. If someone could list the steps and calculations that would be helpful. Thank you!

Answer 1

NaCH3CO2   it is the salt of strong base and weak acid . so anion participate in hydrolysis process

CH3CO2- + H2O ------------------------> CH3COOH + OH-

0.800                                                        0                      0 -------------------> initial

-x                                                              x                         x ---------------------> changed

0.8-x                                                      x                         x -----------------------> equilibrium

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

Kw / Ka = x^2 / 0.8 -x

1.0 x 10^-14 / 1.8 x 10^-5 = x^2 / 0.8 -x

5.56 x 10^-10 = x^2 / 0.8 -x

x^2 + 5.56 x 10^-10 x - 4.45 x 10^-10 = 0

x = 2 .1 x 10^-5

[OH-] = 2 .1 x 10^-5 M

pOH = -log[OH-] = -log (2 .1 x 10^-5)

pOH = 4.68

pH + pOH = 14

pH = 9.32