After adding 0.15 mole of solid NaOH to the solution with a concentration of 0.050 M propanoic acid and 0.080 M sodium propanoate, the pH of the solution is approximately 13.18.
To calculate the pH after adding solid NaOH to the given solutions, we need to consider the reaction between NaOH and propanoic acid (HC₃H₅O₂). This reaction produces sodium propanoate and water.
First, let's determine the initial concentration of propanoic acid (HC₃H₅O₂) and sodium propanoate (C₃H₅O₂⁻) in the solution before adding NaOH.
Given:
- Concentration of propanoic acid (HC₃H₅O₂) = 0.050 M
- Concentration of sodium propanoate (C₃H₅O₂⁻) = 0.080 M
Since propanoic acid (HC₃H₅O₂) is a weak acid, we can assume it dissociates partially. The dissociation equation is:
HC₃H₅O₂ ⇌ H⁺ + C₃H₅O₂⁻
Using the Ka value (1.3 x 10⁻⁵) given for propanoic acid, we can calculate the initial concentrations of H⁺ and C₃H₅O₂⁻:
[H⁺] = [C₃H₅O₂⁻] = √(Ka * [HC₃H₅O₂])
[H⁺] = [C₃H₅O₂⁻] = √(1.3 x 10⁻⁵ * 0.050)
[H⁺] = [C₃H₅O₂⁻] = 1.14 x 10⁻³ M
Now, let's calculate the concentration of OH⁻ ions produced by adding 0.15 mole of NaOH to the solution:
Moles of NaOH = 0.15 mol
Volume of solution = 1.00 L
Concentration of OH⁻ = moles of OH⁻ / volume of solution
Concentration of OH⁻ = 0.15 mol / 1.00 L
Concentration of OH⁻ = 0.15 M
Since NaOH is a strong base, it completely dissociates in water to produce OH⁻ ions. Therefore, the concentration of OH⁻ is equal to 0.15 M.
To determine the pH, we can use the equation:
pH = 14 - pOH
pOH = -log[OH⁻]
pOH = -log(0.15)
pOH ≈ 0.82
pH = 14 - 0.82
pH ≈ 13.18
Therefore, after adding 0.15 mole of solid NaOH to the solution, the pH is approximately 13.18.