Calculate the pH of a solution prepared by dissolving 0.250 mol of benzoic acid (C₇H₅O₂H) and 0.150 mol of sodium benzoate (NaC₇H₅O₂) in water sufficient to yield 1.00 L of solution. The K_aof benzoic acid is 6.50 × 10^-⁵. A) 4.19 B) 2.39 C) 10.0 D) 3.97 E) 4.41

Question

Quizplus · Accepted Answer

First, we need to write out the chemical equation for the dissociation of benzoic acid:
CS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0H ⇌ CS1U1B17S1U1B0HS1U1B15S1U1B0O-S1U1B12S1U1B0 + H+
The equilibrium constant for this reaction is given as KS1U1B1aS1U1B0S1U1B1 S1U1B0= 6.50 × 10S1U1P1-S1S1P0S1U1P1 S1S1P0S1U1P15S1S1P0.

Next, we need to consider the reaction between sodium benzoate and water:
NaCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0 + H2O ⇌ CS1U1B17S1U1B0HS1U1B15S1U1B0O-S1U1B12S1U1B0 + Na+ + OH-
Since sodium benzoate is a salt, it completely dissociates in water. Therefore, we can assume that the [Na+] concentration is equal to 0.150 M.

To calculate the pH, we need to determine the concentrations of H+ and OHS1U1P1-S1S1P0S1U1P1. We can start by using the equilibrium constant for benzoic acid to calculate the concentration of H+:
KS1U1B1aS1U1B0S1U1B1 S1U1B0= [CS1U1B17S1U1B0HS1U1B15S1U1B0O-S1U1B12S1U1B0][H+]/[CS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0]
6.50 × 10S1U1P1-S1S1P0S1U1P1 S1S1P0S1U1P15S1S1P0 = (0.150 mol/L)/(1.00 L) * [H+]/(0.250 mol/L)
Solving for [H+], we get [H+] = 1.167 × 10S1U1P1-S1S1P0S1U1P1 M.

Next, we need to determine the concentration of OHS1U1P1-S1S1P0S1U1P1. Since this is a weak acid-strong base system, we can use the equation:
KW = [OH-][H+]
where KW is the equilibrium constant for water (1.0 × 10S1U1P1-S1S1P0S1U1P1 S1S1P0) and [H+] is the concentration we just calculated. Solving for [OH-], we get [OH-] = 8.59 × 10S1U1P1-S1S1P1S1U1P0 M.

Finally, we can use the equation for pH:
pH = -log([H+])
Substituting in our calculated value for [H+], we get pH = 3.97. Therefore, the answer is D.

Calculate the PH of a Solution Prepared by Dissolving 0