Question

The Ka of benzoic acid (C6H5COOH) is 6.45 x 10-5 . Sodium benzoate (NaC6H5COO) is a common food preservative. If you dissolve 0.100 grams of sodium benzoate into 100.0 mL of water, what will be the pH of this solution?

Answer 1

C6H5COO- + H2O <------> C6H5COOH + OH-

Kb = [C6H5COOH][OH-]/[C6H5COO-][H2O]

Kb = Kw/Ka

Kw = 1×10^-14

Ka = 6.45×10^-5

Kb = 1×10^-14/6.45 × 10^-5

Kb = 1.55 × 10^-10

Kb = [C6H5COOH][OH-]/[C6H5COO-]

at equilibrium

[C6H5COOH] = X

[OH-] = X

Initial concentration of C6H5COONa

Molar mass of C6H5COONa = 144.10g/mole

Mass of C6H5COONa = 0.100g

No of mole = 0.100/144.10= 0.000694 mole

initialConcentration of C6H5COONa = ( 0.000694mole/100ml)×1000ml = 0.00694M

Therefore, [C6H5COO-] at equilibrium = 0.00694- X

1.55×10^-10 = X^2/(0.00694-X)

(0.010757 × 10^-10)- 1.55×10^-10X = X^2

X^2 + (1.55 × 10 ^-10 X) -(0.010757 ×10^-10) = 0

X= 1.037× 10^-6

[C6H5COOH ] = 1.037 ×10^-6M

[OH] = 1.037 ×10^-6M

[C6H5COO-] = 0.00694 - 1.037× 10^-6 = 0.00694M

[OH-] = 1.037 × 10^-6 M

pOH = -log[OH-]

pOH = - log (1.037 × 10^-6)

= 5.98

pH + pOH = 14

pH = 14- 5.98 = 8.02

Therefore, pH of the C6H5COONa solution = 8.02