Question

In a titration of 36.14 mL of 0.3153 M ammonia with 0.4091 M aqueous nitric acid, what is the pH of the solution when 36.14 mL + 10.00 mL of the acid have been added?

Answer 1

Concentration of NH3 = 0.3153 M = 0.3153 mol/L

Volume of NH3 = 36.14 ml = 36.14 L/1000 = 0.03614 L

Number of moles of NH3 = 0.3153 mol/L * 0.03614 L = 0.0114 mol

Concentration of HNO3 = 0.4091 M = 0.4091 mol/L

Volume of HNO3 solution = (36.14 + 10.00) ml = 46.14 ml = 46.14 L/1000 = 0.04614 L

Number of moles of HNO3 = 0.4091 mol/L * 0.04614 L = 0.0189 mol

0.0114 mol of NH3 will react with 0.0114 mol of HNO3 will produce 0.0114 mol of NH4NO3.

Excess moles of HNO3 = (0.0189 - 0.0114) mol = 0.0075 mol

Total volume of solution = (0.04614 + 0.03614) L = 0.08228 L

[HNO3] = 0.0075 mol / 0.08228 L = 0.091 mol/L = 0.091 M

[NH4NO3] = 0.0114 mol/ 0.08228 L = 0.1385 mol/L =0.1385 M

pKb of NH3 = 4.75 so pKa = 14.00 - 4.75 = 9.25

We know from Henderson equation

pH = pKa + log {[conjugate base]/[Acid]}

pH = 9.25 + log (0.1385 / 0.091)

pH = 9.25 + log 1.52

pH = 9.25 + 0.18 = 9.43

pH = 9.43