Question

94. A 50.00-mL solution of 0.0350 M benzoic acid (Ka = 6.4 × 10-5) is titrated with a 0.0209 M solution of sodium hydroxide as the titrant. What is the pH of the acid solution after 15.00 mL of titrant have been added? (Kw = 1.00 × 10-14) a. 1.46 b. 2.83 c. 3.56 d. 10.13 e. 4.19

ANSWER: c

Answer 1

Solution :-

Addition of the NaOH to the benzoic acid converts some of the benzoic acid into the sodium benzoate salt which is the conjugate base therefore the solution becomes buffer of weak acid and its conjugate base

Reaction equation is as follows.

C7H6O2 + OH^- ---- > C7H5O2^- + H2O

Lets calculate the moles of benzoic acid and NaOH

Moles = molarity x volume in liter

Moles of benzoic acid = 0.0350 mol per L * 0.050 L= 0.00175 mol

Moles of NaOH = 0.0209 mol per L * 0.015 L = 0.0003135 mol

After the reaction moles of benzoic acid remain are calculated as

Moles of benzoic acid after reaction = initial moles of benzoic acid – moles of NaOH

                                                                 =0.00175 mol – 0.0003135 mol

                                                                = 0.001437 mol

Moles of conjugate base are same as moles of NaOH that is 0.0003135 mol

The concentration of the acid and conjugate base at the final volume is calculated as follows

Final volume = 50.0 ml + 15.0 ml = 65.0 ml = 0.065L

Molarity = moles / volume

Molarity of benzoic acid = 0.001437 mol / 0.065 L

                                           = 0.0221 M

Molarity of conjugate base = 0.0003135 mol/0.065 L

                                                 = 0.00482 M

Now using the Henderson equation we can find the pH

pH= pka + log [conj base]/[benzoic acid]

pka= -log ka

pka=-log [6.4*10^-5]

      = 4.19

pH= 4.19 + log ([0.00482]/[0.0221])

pH= 3.53

pH is 3.53 which is close to option C

Therefore the answer is option C