Question

Determine the pH of a 0.045M CaSO3. The Ka for SO3 is 1.3x10^-2

Answer 1

ANSWER : pH = 7.27

Consider reaction, SO ₃^2- (aq) + H₂O (l) $\rightleftharpoons$ HSO ₃^- (aq) + OH ^- (aq)

For above reaction, K b = [HSO ₃^- ] [ OH ^- ] / [SO ₃^2- ] = K w / K a = 10 ^-14 / 1.3 $\times$ 10 ^-02 = 7.69 $\times$ 10 ^-13

Let's Use ICE table.

Concentration (M)	SO 32-	HSO 3-	OH -
Initial	0.045
Change	- X	+X	+X
Equilibrium	0.045 - X	X	X

Therefore, K b = [HSO ₃^- ] [ OH ^- ] / [SO ₃^2- ] = (X) (X) / 0.045 - X = 7.69 $\times$ 10 ^-13

X ² / 0.045 - X = 7.69 $\times$ 10 ^-13

Assume X is very small as compared to 0.045. Therefore, we can write 0.045 - X $\simeq$ 0.045.

X ² / 0.045 = 7.69 $\times$ 10 ^-13

X ² = 0.045 $\times$ 7.69 $\times$ 10 ^-13

X ²= 3.4605 $\times$ 10 ^-14

X = 1.8602 $\times$ 10 ^-07 M = [HSO ₃^- ] = [ OH ^- ]

We have relation, [H⁺ ] [ OH ^- ] = 10 ^-14

$\therefore$ [H⁺ ] = 10 ^-14 / [ OH ^- ] = 10 ^-14 / 1.8602 $\times$ 10 ^-07 = 5.376 $\times$ 10 ^-08 M

We have, pH = -log [H⁺ ] = -log 5.376 $\times$ 10 ^-08 = 7.27