Question

Calculate the pH at the equivalence point in titrating .120 M of HClO2 with a solution of .090M of NaOH

Answer 1

HClO₂(aq) + NaOH(aq) ----> NaClO₂(aq) + H2O(l)

At equivalence point all of the HClO₂ is converted to salt ClO₂^-(aq) or NaClO₂(aq)

Since mole ratio between HClO₂(aq) and NaOH(aq) is 1:1

HClO₂(aq):

M1 = 0.120 M

V1 = V L

NaOH:

M2 = 0.090 M

V2 = ??

Since mole ratio between HClO₂(aq) and NaOH(aq) is 1:1, equal moles of each is required i.e

M1*V1 = M2*V2

=> 0.120 M * V L = 0.090 M * V2

=> V2 = (0.120 / 0.090) V L

=> V2 = 1.33V L

Hence final volume of solution, Vt = V + 1.33V = 2.33V L

=> Final concentration of ClO₂^-(aq) = initial moles of ClO₂^-(aq) / Vt = (0.120 * V mol) / 2.33V L

=> [ClO₂^-(aq)] = 0.05143 M

Now ClO₂^-(aq) will undergo hydrolysis as;

--ClO₂^-(aq) + H₂O <------> HClO₂(aq) + OH^-(aq) ; Kb = Kw/Ka = 10^-14 / 1.1*10^-2 = 9.09*10^-13

I: 0.05143 M ------------------- 0 -------------- 0

C: -X ----------------------------- +X ----------- +X

E: (0.05143 - X) M, ------------ X M --------- X M

Kb = 9.09*10^-13 = X² / (0.05143 - X)

Since  X << 0.05143

=> 9.09*10^-13​​​​​​​ = X² / 0.05143

=> X = square root (0.05143 * 9.09*10^-13)

=> X = [OH^-(aq)] = 2.16*10^-7 M

=> pOH = - log( 2.16*10^-7) = 6.665

=> pH = 14 - 6.665 = 7.33 (Answer)