0.1 mol of CH3NH2 (Kb = 5 x 10-4) is mixed with 0.08 mol of HCI and diluted to 1 L. What will be the H+ concentration in the solution?

CH3NH2+H+Cl−⇌CH3NH3+Cl− Initial                  0.1      0.08               - Final                0.02          -              0.08 MCH3NH2+H2O⇌CH3NH3++OH− Kb=CH3NH3+OH−CH3NH2 5×10−4=(0.08)OH−(0.02) OH−=1.25×10−4M H+=KwOH−=1×10−141.25×10−4 =8×10−11M