Question

Consider the following reactions: Koi = 8.3 x 10-8 SO32-(aq) + H2O(1) = HSO:_(aq) + OH(aq) HSO3- (aq) + H2O(1) = HSO3(aq) + OH-(aq) K12 = 7.6 x 10-13 What is the pH of 1.0 x 10-3 M Na2SO3? Select one: O a. 7.0 b. 9.0 C. 5.0 d. 13.0 o e. 10.5

Answer 1

pH = - log[H+] , pOH = -log[OH-], pH = 14-pOH

Thus, to find pH, we need to first find concentration of OH- here.

Na2SO3 will dissociate completely in solution as:

Na2SO3 = 2Na+ + SO32-

So, 1.0 x 10-3 M Na2SO3 solution means a 1 x 10-3 M solution of SO32-.

SO32- reacts with water according to the first equation provided :

SO32- + H2O = OH- + HSO3-

Kb1 =                         […] denotes concentration in molarity

[OH-][H3031 [s032-]

The initial concentration of SO32-is 10-3 M

Let x M SO32-dissociate.

From the dissociation equation, if x M SO32-dissociates, x M OH- and x M HSO3- will be formed.

The ICE table for the reaction is :

	SO32-	OH-	HSO3-
Intial concentration (M)	10-3	0	0
Change (M)	-x	+ x	+x
FInal concentration (M)	10-3 - x	x	x

The HSO3- thus formed will further react with water according to the second equation provided, and produce OH- which will impact the pH. However, the Kb value for the equation is very low, indicating that the amount of reaction occuring, and subsequent OH- produced will be negligible, especially since the starting amount of HSO3- is also low (only a fraction of the HSO3- formed from reaction of SO32- will react).Thus, we neglect contributions to pH from the second reaction.

Now, we need concentration of OH-. Thus, we need value of x.

Putting the concentration values in the equation for Kb1 :

Kb1 = 8.3 x 10-8 =

(x)(x) (0.001-x)

Or, x2 = 8.3 x 10-8(0.001 – x)

Or, x2 + (8.3 x 10-8) x – 8.3 x 10-11 = 0        (Quadratic equation)

Solving,

x = 9.15 x 10-6 (taking only positive root. Use the equation : to solve)

Y = -bvb2-400 2a

Thus, concentration of OH- = x M = 9.07 x 10-6 M

So, pOH = - log[OH-] = 5.04

So, pH = 14 – pOH = 8.96