Question

A 10.0 mL sample of 0.75 M CH3CH2COOH(aq) is titrated with 18.0 mL of 0.30 M NaOH(aq) (adding NaOH to CH3CH2COOH). Determine which region on the titration curve the mixture produced is in, and the pH of the mixture. Ka of CH3CH2COOH is 1.3 \times × 10−5.

Henderson–Hasselbalch equation: LaTeX: pH=pK_a+\log\frac{\left[base\right]}{\left[acid\right]} p H = p K a + log ⁡ [ b a s e ] [ a c i d ]

2046 HW10 Q1-1.jpg

1) After adding the NaOH solution, the mixture is (before, after, at) the equivalence point on the titration curve.

2) The pH of the solution after adding NaOH is (4.48, 5.30, 8.70, 7.00) .

Answer 1

Q1. After adding the NaOH solution, the mixture is before the equivalence point on the titration curve.

Q2. The pH of the solution after adding NaOH is 5.30

Explanation

moles acetic acid = (molarity acetic acid) * (volume acetic acid)

moles acetic acid = (0.75 M) * (10.0 mL)

moles acetic acid = 7.5 mmol

moles NaOH = (molarity NaOH) * (volume NaOH)

moles NaOH = (0.30 M) * (18.0 mL)

moles NaOH = 5.4 mmol

Since moles NaOH added is less than moles acetic acid present, therefore the mixture has not reached equivalence point because at equivalence point, moles acid = moles base

pH is calculated Henderson-Hasselbalch equation

initial moles acetic acid = 7.5 mmol

moles NaOH added = 5.4 mmol

moles acetic acid remaining = initial moles acetic acid - moles NaOH added

moles acetic acid remaining = 7.5 mmol - 5.4 mmol

moles acetic acid remaining = 2.1 mmol

moles conjugate base formed = moles NaOH added = 5.4 mmol

Ka = 1.3 x 10^-5

pKa = -log(Ka)

pKa = -log(1.3 x 10^-5)

pKa = 4.89

According to Henderson - Hasselbalch equation

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

pH = pKa + log(moles conjugate base / moles acetic acid remaining)

pH = 4.89 + log(5.4 mmol / 2.1 mmol)

pH = 5.30