Question

a 500.00 mL solution is prepared by dissoving 6.20 sodium acetate in water. the pKa of the conjugate acid (acetic acid) is 4.75. Calculate the solution pH and the concentration of acetate and acetic acid (in molarity mol/L). Use the molecular weight of sodium acetate of 82.0 g/mol.

Answer 1

MOLARITY OF CH3COONa = (w/M)*(1000/V in ml)

                      = (6.2/82)*(1000/500)

                      = 0.15 M

pkb= 14-4.75 = 9.25

kb = 10^-9.25 = 5.623*10^-10

CH3COONa(aq) ----> CH3COO^-(aq) + Na^+(aq)

           CH3COO^-(aq) + H2O(l) ----> CH3COOH(aq) + OH^-(aq)

initial     0.15 M                        0 M         0 M

chnage       -x                         +x M          +x M

Equil       0.15-x M                      x            x M

   Kb = [OH-][CH3COO-]/[CH3COOH]

(5.623*10^-10) = x*x/(0.15-x)

x =9.18*10^-6

[OH-]= X = 9.18*10^-6 M

[CH3COO-] = X = 9.18*10^-6 M

[CH3COOH] = 0.15-X = 0.15 -(9.18*10^-6) = 0.14999 M

pH = 14 - (-log(OH-))

     = 14 - (-log( 9.18*10^-6))

    = 8.96