Answer to Question #128772 in General Chemistry for Jocabed Jimenez

Question

The pH of a weak base 0.10 M is 12.30 What is the Kb of the base and its percent ionization?


What is the H+ of a solution of pyridine, C5H5N, 0.0020 M? Kb = 1.7 x10-9

Calculate the pH of a 0.015 M NaOCl solution and the % hydrolysis of this salt. Ka HOCl = 2.9 x 10-8


One liter of 0.400 M NH3 solution also contains 12.78 g of NH4Cl. How much will the pH of the solution change if 0.075 mol of gaseous HCl is bubbled into it? KbNH3 = 1.8 x 10-5
pH changes from ___________________ to ____________________

Answer 1

1. pН = 14 – 0.5рК_В + 0.5lgС

0.5рК_В=13.5-12.3

рК_В=2.4

К_В = 3.98x10^-3

ionization percent = [OH^-]/[Base]= (10^12.3-14)/0.1=20%

2. pН = 14 – 0.5рК_В + 0.5lgС = 8.26

[H+]=10^-8.26=5.5x10^-9

3. [OH^—] = (K_hydrolysis·[NaOCl])^1/2

K_hydrolysis=K_w/K_a=10^-14/2.9x10^-8=3.45x10^-7

[OH^—]=7.2x10^-5

pH=14-p[OH]=9.86

4. [NH₄Cl] = 12.78/53.5=0.24 M

pН₁= 14 - рК_b + lg [NH₃]/[NH₄Cl] = 9.255 + lg(0.4/0.24) = 9.477

pН₂ = 14 - рК_b + lg [NH₃]/[NH₄Cl] = 9.255 + lg(0.4-0.075/0.24+0.075) = 9.268

pH changes from 9.477 to 9.268