The pH of a solution that contains 0.818 M acetic acid (Ka = 1.76 x 10-5) and 0.172 M sodium acetate is
__________. The Ka of acetic acid is 1.76 × 10-5.
Use the Ka expression (or the Henderson-Hasselbalch expression).
Ka = (H^+)(CH3COO^-)/(CH3COOH)
You are given CH3COO^- and CH3COOH, solve for H^+, then convert to pH.
To determine the pH of the solution, you can use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa and the ratio of the concentrations of the conjugate acid and base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([conjugate base]/[acid])
In this case, acetic acid (CH3COOH) is the acid and sodium acetate (CH3COONa) is the conjugate base. The pKa value of acetic acid is given as 1.76 x 10^-5.
First, we need to calculate the ratio of the concentrations of the conjugate base and acid:
[conjugate base] / [acid] = [CH3COO-] / [CH3COOH]
Given the molarities of the solutions, we have:
[conjugate base] = 0.172 M (concentration of sodium acetate)
[acid] = 0.818 M (concentration of acetic acid)
Now, let's substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log([0.172] / [0.818])
Using the given pKa = 1.76 x 10^-5, and performing the calculations:
pH = -log(1.76 x 10^-5) + log(0.172 / 0.818)
pH = -(-4.755) + log(0.210)
pH = 4.755 + (-0.677)
pH = 4.078
Therefore, the pH of the solution is approximately 4.078.