Question

A 23.74-mL volume of 0.0981 M NaOH was used to titrate 25.0 mL of a weak monoprotic acid solution to the stoi- chiometric point. Determine the molar concentration of the weak acid solution. Express your answer to the correct number of significant figures.

Answer 1

Answer:

Molar concentration of the weak acid solution is 0.0932

Explanation:

Using the formula: $\frac{C_aV_a}{C_bV+b} = \frac{n_a}{n_b}$

Where Ca = molarity of acid

Cb = molarity of base = 0.0981 M

Va = volume of acid = 25.0 mL

Vb = volume of base = 23.74 mL

na = mole of acid

nb = mole of base

Since the acid is monopromatic, 1 mole of the acid will require 1 mole of NaOH. Hence, na = nb = 1

Therefore, $C_a = \frac{C_bV_b}{V_a}$

Ca = 0.0981 x 23.74/25.0

                 = 0.093155 M

To 4 significant figure = 0.0932 M

Answer 2

Explanation:

Given:

Vb = 23.74 mL

Cb = 0.0981 M

Va = 25.0 mL

Since the weak acid is monoprotonic. Therefore, 1 mole of NaOH reacts with 1 mole of acid. By stoichiometry, the number of moles of acid = molar conc. × volume

= 0.0981 × 23.74 × 10^-3

= 0.00233 mole.

Molar concentration of acid = 0.00233/25 × 10^-3

= 0.093 M.