Question

1.) Which of the following will be more soluble in an acidic solution than in pure water?

SrSO₄, PbCl₂, RbClO₄, CuCN, Be(OH)₂

2.) Calculate the pH at the equivalence point for the titration of 0.240 M methylamine (CH3NH2) with 0.240 M HCl. The Kb of methylamine is 5.0× 10–4.

Answer 1

1) Be(OH)2 is basic compound so it is more soluble in acidic solution.

2) CH3NH2 + H+ ---> CH3NH3+

Suppose 1 L of CH3NH2
To reach the equivalence point since the concentrations of HCl and CH3NH2 are the same we need 1 L of HCl
Total volume becomes 2 L
At the equilvalence point 0.240 mol CH3NH2 + 0.240 mole H+ ----> 0.240 mole CH3NH3+

[CH3NH3+] = 0.24/ 2 L = 0.12M

CH3NH3+ <-----> CH3NH2 + H+

K = Kw/Kb = 1.0 x 10^-14 / 5.0 x 10^-4 = 2.0 x 10^-11

2.0 x 10^-11 = x^2 / 0.12-x      ( as x is too small then 0.12 -x = 0.12)

x = [H+]= 1.55 x 10^-6 M

pH = 5.81