ionization constant for acid = Ka
let's assume that monoprotic acid is HA
the reaction between HA and NaOH,
HA(aq) + NaOH(aq) → NaA(aq) + H₂O(l)
the stoichiometric ratio between HA and NaOH is 1 : 1
Hence,
Moles of NaOH added = reacted moles of HA in 25.00 mL
Moles of NaOH added = concentarion x volume added
= 0.104 mol L⁻¹ x 21.70 x 10⁻³ L
reacted moles of HA in 25.00 mL = 0.104 mol L⁻¹ x 21.70 x 10⁻³ L
hence, the initial [HA] = moles / volume
= (0.104 mol L⁻¹ x 21.70 x 10⁻³ L) / 25.00 x 10⁻³ L
= 0.090 M
According to the pH, the molar solubility = [H⁺(aq)] = X
pH = -log[H⁺(aq)]
3.70 = -log[H⁺(aq)]
[H⁺(aq)] = 1.995 x 10⁻⁴ M
X = 1.995 x 10⁻⁴ M
at equilibrium,
HA(aq) ⇄ H⁺(aq) + A⁻(aq)
Initial 0.090
Change -X +X +X
Equilibrium 0.090 - X X X
Ka = [H⁺(aq)] [ A⁻(aq)] / [HA(aq)]
= X x X / (0.090 - X)
= (1.995 x 10⁻⁴ M)² / (0.090 - 1.995 x 10⁻⁴) M
= 4.432 x 10⁻⁷ M