Question

Determine the pH of a 0.045 M    CaSO3.    

The Ka for H2SO3 is 1.3 * 10 ^   (-2)

Answer 1

Consider the dissociation of CaSO₃ as below:

CaSO₃ (aq) <=====> Ca²⁺ (aq) + SO₃^2- (aq)

The dissociation of H₂SO₃ can be shown as below:

H₂SO₃ (aq) ------> H⁺ (aq) + HSO₃^- (aq); K_a1 = 1.3*10^-2

HSO₃^- (aq) ------> H⁺ (aq) + SO₃^2- (aq); K_a2 = 6.6*10^-8

CaSO₃ dissociates to SO₃^2- which establishes equilibrium as below:

SO₃^2- (aq) + H₂O (l) <=====> HSO₃^- (aq) + OH^- (aq)

Since OH^- is formed, we must work with K_b; given K_a2, we can find out K_b2.

K_b2 = K_w/K_a2 = (1.0*10^-14)/(6.6*10^-8) = 1.5151*10^-7

Set up the expression for K_b.

K_b = [HSO₃^-][OH^-]/[SO₃^2-] = (x).(x)/(0.045 – x)

===> 1.5151*10^-7 = x²/(0.045 – x)

Make an assumption; since K_b is extremely small, x is small compared to 0.045 M. Therefore,

1.5151*10^-7 = x²/0.045

====> x² = 6.81795*10^-9

====> x = 8.257*10^-5

Therefore, [OH^-] = 8.257*10^-5 M; hence pOH = -log [OH^-] = -log (8.257*10^-5) = 4.083 ≈ 4.08

Since pH + pOH = 14, therefore, pH = 14 – Poh = 14 – 4.08 = 9.92 (ans).