Answer:
Molar concentration of the weak acid solution is 0.0932
Explanation:
Using the formula: [tex]\frac{C_aV_a}{C_bV+b} = \frac{n_a}{n_b}[/tex]
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, [tex]C_a = \frac{C_bV_b}{V_a}[/tex]
Ca = 0.0981 x 23.74/25.0
= 0.093155 M
To 4 significant figure = 0.0932 M
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.
