Determine the pH of a solution.

Question

Determine the pH of the solution after the addition of 549 mL of 0.158 M strontium hydroxide (Sr(OH)₂) to 786 mL of 0.919 M benzoic acid (C₆H₅COOH).

Assume that the volumes are additive and that the 5% approximation is valid.

The pK_a for benzoic acid is 4.20

Answer 1

2C₆H₅COOH + Sr(OH)₂ ==> 2H₂O + Sr(C₆H₅COO)₂ .. balanced equation

initial moles benzoic acid = 786 ml x 1 L / 1000 ml x 0.919 mol / L = 0.722 mols

initial moles Sr(OH)₂ = 549 ml x 1 L / 1000 ml x 0.158 mol / L = 0.0867 mols

final volume = 549 ml + 786 ml = 1335 ml = 1.335 L

Set up ICE table:

2C₆H₅COOH + Sr(OH)₂ ==> 2H₂O + Sr(C₆H₅COO)₂

0.722..............0.0867.....................................0................Initial

-0.173............-0.0867..................................+0.0867.......Change

0.549...................0........................................0.0867........Equilibrium

Final [C₆H₅COOH] = 0.549 mol / 1.335 L = 0.411 M

Final [C₆H₅COO-] = 0.0867 mol x 2 / 1.335 L = 0.130 M

Since we have a weak acid and the conjugate base, we have a buffer and we can use the Henderson Hasselbalch equation to find the pH.

Henderson Hasselbalch equation: pH = pKa + log [C₆H₅COO-] / [C₆H₅COOH]

pH = 4.20 + log (0.130 / 0.411)

pH = 4.20 + log 0.316 = 4.20 - 0.500

pH = 3.700