Question

0.850 mol of a weak acid, HA, and 12 g of NaOH are placed in enough water to produce 1.00 L of solution. The final pH of the solution produced is 5.2. Calculate the ionization constant, Kb, of A- (aq).

Answer 1

[HA] = 0.850 mol/ 1 L = 0.85 M

Molar mass of NaOH = 40 g/mol

No of moles of NaOH = 12 g / 40 g/mol = 0.3 mol

[NaOH] = 0.3 mol/1L = 0.3 M = [OH-]

Step 1: write the net ionic equation

Step 2 : make ICE table

Step 3 : use the Henderson–Hasselbalch equation.

pH = pKa + log [base] /[acid]

Step 4 : calculate pKa

Step 5 : calculate pKb from pKa by using the formula,

pKa + pKb = 14

The net ionic equation is -

Answer : 2.91 × 10^-9

** The process is absolutely fine. If you don't understand, comment below.