Question

This question has several parts to it.
The givens for the problem are : o-Toluic acid has a molecular weight of 136.15 and the pKa of o-toluic acid is 3.90

a) How many moles of toluic acid are present in 0.134 g?

​b)What is the pH of a solution of 0.134 g of o-toluic acid in 200 mL of water?

c) If 0.134 g of o-toluic acid is dissolved in water and the solution is titrated with 0.250 M NaOH, what volume of titrant is needed to reach the equivalent point?

​d) What is the pH at the equivalence point?

​Please help me understand this stuff.

Answer 1

a ) moles = mass / molar mass

                = 0.134 / 136.15

                = 9.84 x 10^-4

b) molarity of acid = moles / volume (L)

                            = 9.84 x 10^-4 / 0.2

                           = 4.92 x 10^-3 M

pKa = 3.90

pH = 1/2 [pKa -logC]

pH = 1/2 [3.90 -log (4.92 x 10^-3)]

pH = 3.10

c)

at equivalece point moles of acid = moles of base

9.84 x 10^-4   = 0.250 x V

V = 3.92 x 10^-3 L

volume of base needed = 3.92 mL

d) at equivlaece point only salt remains

salt concentration = 9.84 x 10^-4 x 1000 / (200 + 3.92 ) = 4.82 x 10^-3 M

pH = 7 + 1/2 [pKa + log C]

pH = 7 + 1/ 2[3.90 + log 4.82 x 10^-3]

pH = 7.79