Na H3O OH C6H5CO2Solutionmoles benzoic acid 0235 g 12212 g

Na⁺, H₃O⁺, OH^-
C₆H₅CO₂^-

moles benzoic acid = 0.235 g / 122.12 g/mol=0.00192

moles NaOH required = 0.00192

volume of NaOH required = 0.00192

volume of NaOH = 0.00192 / 0.108 M=0.0178 L

total volume = 0.0178 L + 0.100 L = 0.118 L

moles benzoate formed = 0.00192

concentration benzoate = 0.00192 / 0.118 L=0.0163 M

C6H5COO- + H2O <=> C6H5COOH + OH-

Kb = Kw/Ka = 1.0 x 10^-14 / 6.3 x 10^-5 = 1.6 x 10^-10 M

1.6 x 10^-10 = x^2 / 0.0163-x

x = [OH-]= [C6H5COOH]= 1.61 x 10^-6 M

[H+]= 1.0 x 10^-14 / 1.61 x 10^-6=6.21 x 10^-9

pH = 8.21

[Na+]= 0.00192 / 0.118 L=0.0163 M