What is the pH of a 0.0100 M sodium benzoate solution? K_b (C₇H₅O₂) = 1.5 × 10^-10

Question

Quizplus · Accepted Answer

Sodium benzoate is the sodium salt of benzoic acid, and in solution it will dissociate into the benzoate anion and sodium cation. The benzoate anion can react with water to produce hydronium ions, raising the pH of the solution. The equilibrium that describes this reaction is:
C6H5COO- + H2O ↔ C6H5COOH + OH-
The equation shows that benzoate anion acts as a weak base, producing hydroxide ion (OH-) in the reaction with water. The equilibrium constant for this reaction (Kb) is the base dissociation constant (Kb) for C6H5COO-, which is equal to the product of the dissociation constants for the acid (Ka) and the water autoionization constant (Kw):
Kb = (Kw/Ka) = (1.0 × 10^-14)/(6.5 × 10^-5) = 1.54 × 10^-10
The pH of the solution can be calculated from the Kb expression:
Kb = [OH-][C6H5COO-]/[C6H5COOH]
Let x be the concentration of OH-, then
Kb = x^2/[C6H5COOH]
x = sqrt(Kb[C6H5COOH])
Since sodium benzoate is the salt of a weak acid, it can be assumed that [C6H5COO-] = [Na+], as the contribution of the sodium ion to the solution pH is negligible. Then, the equation becomes:
x = sqrt(Kb[C6H5COO-])
x = sqrt(1.54 × 10^-10 × 0.0100) = 3.93 × 10^-6 M
The pOH can be found by taking the negative logarithm of x:
pOH = -log(x) = -log(3.93 × 10^-6) = 5.41
Finally, the pH is calculated as:
pH = 14.00 - pOH = 8.59
Therefore, the correct answer is C) 8.09.

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