Question

The pK of saccharin, HC₇H₃SO₃, a sweetening agent, is 11.68.

What is the pKb of the saccharinate ion, C7H3SO3-?

Does a solution of sodium saccharinate in water have a 
pH of 7, or is the solution acidic or basic?

If the pH is not 7, calculate the pH of a 0.010 M solution of sodium saccharinate in water.

Answer 1

a)

if pKa of saccharin is = 11.68

then

pKb = 14 - pKa = 14-11.68= 2.32

pKb = 2.32

b)

NO, since the sodium saccharinate will enter solution as

NaC7H3SO3 = Na+ C7H3SO3-

C7H3SO3- + H2O = HC7H3SO3 + OH-

Then, expect this to be basic

c)

pH if

M = 0.01

then

C7H3SO3- + H2O = HC7H3SO3 + OH-

Kb = [HC7H3SO3][OH-]/[C7H3SO3-]

Kb = 10^-pKb = 10^-2.32 = 0.00478

0.00478 = (x*x)/(0.01-x)

solve for x

x = 0.004925

pOH = -log(OH) = -loG(0.004925) = 2.30751

pH = 14-2.30751 = 11.69249