Question

Part Calculate the pH of a solution prepared by mixing 15.0 mL of 0.10 M NaOH and 30.0 mL of 0.10 M benzoic acid solution. (Benzoic acid is monoprotic; its dissociation constant is 6.5x10.) IVO AQ R O a ? Submit Request Answer

Answer 1

Part C : pH = 4.19

Explanation

concentration benzoic acid = 0.10 M

volume benzoic acid = 30.0 mL

moles benzoic acid = (concentration benzoic acid) * (volume benzoic acid)

moles benzoic acid = (0.10 M) * (30.0 mL)

moles benzoic acid = 3.0 mmol

Similarly, moles NaOH = 1.5 mmol

moles conjugate base formed = moles NaOH added

moles conjugate base formed = 1.5 mmol

moles benzoic acid remaining = (initial moles benzoic acid) - (moles conjugate base formed)

moles benzoic acid remaining = (3.0 mmol) - (1.5 mmol)

moles benzoic acid remaining = 1.5 mmol

K_a benzoic acid = 6.5 x 10^-5

pK_a = -log(K_a)

pK_a = -log(6.5 x 10^-5)

pK_a = 4.19

According to Henderson - Hasselbalch equation,

pH = pK_a + log([conjugate base] / [weak acid])

pH = pK_a + log(moles conjugate base formed / moles benzoic acid remaining)

pH = 4.19 + log(1.5 mmol / 1.5 mmol)

pH = 4.19 + log(1)

pH = 4.19 + 0

pH = 4.19