340 ml of a 0150 M solution of NH3 aq is titrated with 0100

340 ml of a 0.150 M solution of NH3 (aq) is titrated with 0.100 M HCl. Calculate the pH fo the solution after 750 ml of the HCl has been added. (Ka (NH4+) = 5.6 x 10 -10)

Solution

Moles of NH3 = 0.15 x 0.34 = 0.051

Moles of HCl = 0.1 x 0.75 = 0.075

Moles of NH4+ formed = 0.051 = 0.051 / 1.09 = 0.0468 M

Moles of HCl left = 0.075 - 0.051 = 0.024 moles = 0.024 / 1.09 = 0.022 M

NH4+ + H2O --------> NH3 + H3O+ (or H+)

Ka = X * (0.022 + X) / (0.0468 - X) = 5.6 x 10^-10

=> X = 1.19 x 10^-9 M

[H+] = 0.022 + X = 0.022

pH = - log [H+] = 1.658