Question

For the titration of 25 ml of a 0.025 M weak acid solution (pKa = 4.615) with 0.050 M NaOH. What will be the pH and pOH at the Equivlance point?

Answer 1

pKa = 4.615

pKa = -logKa

Ka = antilog (-4.615) = 2.43 *1 0^-5

Now, K_b = Kw/Ka = 10^-14/(2.43 *1 0^-5) = 4.12 * 10^-10

Molarity of weak acid (M₁) = 0.025 M

Volume of weak acid (V₁) = 25 mL = 0.025 L

Molarity ofNaOH (M₂) = 0.050M

Let volume of NaOH be V₂

From Molarity equation,

M₁V₁ = M₂V₂

0.025 M * 25 mL = 0.050 mL *V₂

_V2 = 12.5 mL

Total volume of solution = 25 +12.5 = 37.5 mL

Thus, in solution concentration of weak acid([HA])= (0.025M*25mL)/37.5mL

= 0.017M =[NaOH] = [OH^-]

At equivalence point, we have A^- which will reacts with water as:

A^- + H₂O    HA + OH^-

I(M) 0.17 0 0

C(M) -x +x +x

E(M) 0.17-x x x

K_b = [ HA][OH^-]/[A^-]

4.12 * 10^-10 = x.x / 0.17-x = x²/ (0.17-x)

(7.0* 10^-11) -(4.12 * 10^-10)x = x²

x = 8.37 * 10^-6 = [OH^-]

pOH = -log[OH^-] = -log( 8.37 * 10^-6 ) = 5.08

so, pH = 14 -pOH = 14 -5.08 = 8.92