Question

Determine the volume (in mL) of 0.337 M sodium hydroxide (NaOH) that must be added to 945 mL of 0.0699 M ascorbic acid (C₅H₇O₄COOH) to yield a pH of 4.72. 

Assume the 5% approximation is valid and report your answer to 3 significant figures. A table of pKa values can be found here.

Answer 1

We have Henderson eq

pH = pka + log [ conjugate base] / [ acid]

pka is 4.1 , pH needed = 4.72

hence by eq

4.1 = 4.72 + log [ C5H7O4COO-] /[ C5H7O4COOH]

[C5H7O4COO-] / [C5H7O4COOH] = 10 ^ ( 4.1-4.72) = 0.24

[C5H7O4COO-] = 0.24 [C6H&O4COOH] ( Molarity = moles / vol , vol is same for both , hence cancells)

C5H7O4COO- moles = 0.24 C5H7O4COOH moles ..............(1)

Let NaOH volume be V liters , NaOH moles = M x V = 0.337V

we have reaction

C5H7O4COOH (aq) + OH -(aq) <--> C5H7O4COO-(aq) + H2O (l)

Initially C5H7O4COOH moles = M x V = 0.0699 x ( 0.945) = 0.066

NaOH moles added = C5H7O4COO- moles = 0.337V

C5H7O4COOH moles now = 0.066-0.337V   

now substituting above in eq1

( 0.066-0.337V) = 0.24 ( 0.337V)

0.066-0.337V = 0.0809 V

V = 0.158 L = 158 ml

Thus volume of NaOH needed = 158 ml