A solution of an unknown acid had a ph of 3.70. titration of a 25.0 ml aliquot of the acid solution required 21.7 ml of 0.104 m sodium hydroxide for complete reaction. assuming that the acid is monoprotic, what is its ionization constant

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