Question

A 0.900 g sample of phthalic acid re- quired 40.0 mL of NaOH solution for complete titration. What weight ofben- zoic acid would require 25.0 mL of the same NaOH solution for the titration? In the formula for benzoic acid shown to the right, the acidic hydrogen is circled . Complete and balance the following acid-base reactions:

Answer 1

Formula for phthalic acid: HOOC(C₆H₄)COOH

HOOC(C₆H₄)COOH(aq)+2NaOH(aq)
→(OOC(C₆H₄)COO)Na₂(aq)+2H₂O(l)

We shall first determine the number of moles of phthalic acid used.

Relative Molecular Mass(RMM) of phthalic acid
= 6(RAM of H)+4(RAM of O)+8(RAM of C)
= 6(1)+4(16)+8(12)
= 166

166g= 1 mole of phthalic acid
0.900g= 0.900/166
= 5.42x10⁻ ³ moles of phthalic acid

According to the equation, the ratio of the number of moles of phthalic acid needed to that of NaOH needed is 1:2. Therefore, the number of moles of NaOH needed
= 2(number of moles of phthalic acid)
= 2(5.42x10⁻ ³)
= 0.0108

[NaOH] needed
= number of moles of NaOH/volume of NaOH solution in L
= 0.0108/(40.0x10⁻ ³)
= 0.271M

Number of moles of NaOH in 25.0mL solution of 0.271M NaOH
= ([NaOH])x(volume of the solution in L)
= (0.271)x(25.0x10⁻ ³)
= 6.777x10⁻ ³

Formula for benzoic acid: C₆H₅COOH

C₆H₅COOH(aq)+NaOH(aq)
→C₆H₅COONa(aq)+H₂O(l)

According to the equation, the ratio of the number of moles of benzoic acid needed to that of NaOH needed is 1:1. Therefore, the number of moles of C₆H₅COOH needed
= number of moles of NaOH needed
= 6.777x10⁻ ³

Relative Molecular Mass(RMM) of benzoic acid
= 7(RAM of C)+6(RAM of H)+2(RAM of O)
= 7(12)+6(1)+2(16)
= 122

1 mole of benzoic acid= 122g
6.777x10⁻ ³ moles of benzoic acid
= (122)(6.777x10⁻ ³)
= 0.827g

Thus, 0.827g of benzoic acid are required.