Question 6.11: Calculate the pH of 0.050 M NH3. State any assumptions made ...

Since NH_3 is a weak base ( K_b = 1.75 × 10^{–5}), we assume that

[OH^–] >> [H_3O^+]           and          C_{NH3} = 0.050 M

With these assumptions, we find (be sure to check the derivation)

[OH^–] = \sqrt{K_bC_{NH_3}} = \sqrt{(1.75\times 10^{-5})(0.050)} =9.35 \times 10^{-4}  M

Both assumptions are acceptable (again, verify that this is true). The concentration of H_3O^+ is calculated using K_{\mathrm{w}}

[H_3O]= \frac{K_{\mathrm{w}}}{[OH^-]} =\frac{1.00 \times 10^{-14}}{9.35 \times 10^{-4}} =1.07 \times 10^{-11}

giving a pH of 10.97.