What is the pH of a solution that is 0.40 M in sulfurous acid (H2SO3) and )0.22 M in sodium sulfite (NaHSO3-)? Ka (H2SO3) = 1.3x10-2
Use the Henderson-Hasselbalch equation.
pH = pKa + log (base)/(acid).
Thanks :) I did that and got 1.63 as my pH, which is the correct answer.
To determine the pH of a solution containing a weak acid and its conjugate base, you need to use the equation for the acid dissociation constant (Ka) and apply the principles of the Henderson-Hasselbalch equation.
Here's how to get the answer:
Step 1: Write the balanced chemical equation for the dissociation of sulfurous acid (H2SO3) in water:
H2SO3 ⇌ H+ + HSO3-
Step 2: Set up the expression for the acid dissociation constant (Ka) using the concentrations of the acid (H2SO3) and its conjugate base (HSO3-):
Ka = [H+][HSO3-] / [H2SO3]
Step 3: Substitute the given values into the equation:
Ka = [H+][HSO3-] / (0.40 M)
Step 4: Rearrange the equation to solve for [H+]:
[H+] = (Ka * [H2SO3]) / [HSO3-]
Step 5: Substitute the values into the equation:
[H+] = (1.3 x 10^-2) * (0.40 M) / (0.22 M)
Step 6: Calculate [H+]:
[H+] = 0.024
Step 7: Finally, calculate the pH of the solution:
pH = -log[H+]
pH = -log(0.024)
pH ≈ 1.62
Therefore, the pH of the solution is approximately 1.62.