Question

The titration curve for a 25.0 ml sample of benzoic acid shows that 10.0 ml of 0.100M NaOH is required to reach the equivalence point. (See #3 for Ka info.) i) What is the concentration of the original benzoic acid sample? ii) What is the pH at the equivalence point? iii) Calculate the pH after addition of 5.00 ml of NaOH and the pOH after addition of 20.0 ml NaOH.

Answer 1

i. C1V1 = C2V2

C1 * 25 ml = 0.1M * 10 ml

C1 = 0.04M

ii. 1. We know the following:

C6H5COOH + OH¯ --->C6H5COO¯ + H₂O

10.0 mL of NaOH is required to reach equivalence.

The total volume of the solution is 35.0 mL.

The moles of sodium benzoate are 0.001 mol

2) Calculate the molarity of the sodium benzoate:

0.001 mol / 0.035 L = 0.0286 M

3) Calculate the K_b of sodium benzoate:

K_w = K_aK_b

1.00 x 10^-14 = (6.4 x 10^-5 ) (x)

x = 1.5625 x 10^-10

4) Calculate pH of the solution:

1.5625 x 10^-10 = [(x) (x)] / 0.0286

x = 2.114 x 10^-6 M (this is the hydroxide ion concentration)

pOH = 5.675

pH = 8.325