Question

Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8×10−5) with 0.20 M HNO3. Calculate the pH after addition of 50.0 mL of the titrant at 25 ∘C.

Answer 1

The titration takes place between NHI Volume of 0.20 M NH, is 50.0 mL Volume of the titrant added is 50.0 mL Since the total volume is doubled, concentration will be halved. Thus, Initial concentration of NH4+- Write the ICE table for the following reaction as shown below. 0.20 M 2 r14 = 0.01 M

0.10 M 00 Initial: Change: Equilibrium: 0.10-x x x The expression of Ka for the reaction can be written as shwon below. NH4+ Kb value of NH, is 1.8 x 10 Kw 1.0×10-14 Kb 1.8×10-5 Ka-- = 0.555 x 10-9 Substitute the the value of Ka and the equilibrium concentrations in the expression.

0.555×10-9 (%) 0.10-x 0.555 x 109- " 7.45 × 10-6 x From the ICE table, Concentration of H.| H--х So, | H- 1= 7.45 × 10-6 M Calculate pH of the solution as sshown below. pH-log[H] ー -log(7.45 × 10-6 = 5.13 Therefore, pH of the solution is 5.13