Question

what is the pH of a 0.350M MgF2 solution? Ka of HF = 3.5x10^-4

Answer 1

Sol :-

Given,

Concentration of MgF₂ = 0.350 M

Because, MgF₂ is strong electrolyte, therefore on complete dissociation it gives 1 mol of Mg²⁺ ion and 2 moles of F^- ions as :

MgF₂ ----------------------> Mg²⁺ + 2F^-

So,

Concentration of F^- = 2 x Concentration of MgF₂ = 2 x 0.350 M = 0.700 M

Also given,

Acid dissociation constant of HF = k_a = 3.5 x 10^-4

So,

Base dissociation constant of F^- = k_b = k_w (Ionic product of water)/k_a

k_b = 1.0 x 10^-14 / 3.5 x 10^-4

k_b = 2.86 x 10^-11  

ICE table of F^- is :

................................F^-..................+.................H₂O <-----------------------> HF................+...............OH^-

Initial (I)..................0.700 M.....................................................................0.0 M.............................0.0 M

Change (C)..............-y...............................................................................+y..................................+y

Equilibrium (E)......(0.700-y) M....................................................................y M................................y M

y = Amount dissociated per mole

Expression of k_b is :

k_b = [HF].[OH^-] / [F^-]

2.86 x 10^-11 = y²/((0.700-y)

y<<<0.700, so neglect y as compare to 0.700

y² = 2.86 x 10^-11 x 0.700

y = 4.47 x 10^-6

So,

[OH^-] = y M = 4.47 x 10^-6  M

Now,

pOH = - log [OH^-]

So,

pOH = - log 4.47 x 10^-6 = 5.35

Also,

pH + pOH = 14

So,

pH = 14-pOH

pH = 14 - 5.35

pH = 8.65

Hence, pH = 8.65