Question 18.11: Determining pH from Kb and Initial [B] Problem Dimethylamine......

Plan We know the initial concentration (1.5 M) and K_b  (5.9×10^{−4}) of (CH_3)_2NH and have to find the pH. The amine reacts with water to form OH^{−}, so we have to find [OH^−] and then calculate [H_3O^+] and pH. We first write the balanced equation and K_b expression. Because K_b  >>  K_w, the [OH^−] from the autoionization of water is negligible, so we disregard it and assume that all the [OH^−] comes from the base reacting with water. Because K_b is small, we assume that the amount of amine reacting, [(CH_3)_2NH]_{\text{reacting}}, can be neglected. We set up a reaction table, make the assumption, and solve for x. Then we check the assumption and convert [OH^−] to [H_3O^+] using K_w; finally, we calculate pH.

Solution Writing the balanced equation and K_b expression:
            (CH_3)_2NH(aq)  +  H_2O(l)  \xrightleftharpoons[]{}  (CH_3)_2NH_2^+(aq)  +  OH^−(aq)
            K_b  =  \frac{[(CH_3)_2NH_2^+][OH^−]}{[(CH_3)_2NH]}

Setting up the reaction table (Table 1), with x  =  [(CH_3)_2NH]_{\text{reacting}}  =  [(CH_3)_2NH_2^+]  =  [OH^−] :

Making the assumption: K_b is small, so
            [(CH_3)_2NH]_{\text{init}}  −  [(CH_3)_2NH]_{\text{reacting}}  =  [(CH_3)_2NH]  ≈  [(CH_3)_2NH]_{\text{init}}
Thus, 1.5 M − x ≈ 1.5 M.
Substituting into the K_b expression and solving for x:

           K_b  =  \frac{[(CH_3)_2NH_2^+][OH^−]}{[(CH_3)_2NH]}  =  5.9×10^{−4}  ≈  \frac{x^2}{1.5}
           x  =  [OH^−]  ≈  3.0×10^{−2}  M

Checking the assumption:
           \frac{3.0×10^{−2}  M}{1.5  M}  ×  100  =  2.0\% (< 5%; assumption is justified.)
Note that the Comment in Sample Problem 18.8 applies in these cases as well:

Calculating pH:
          [H_3O^+]  =  \frac{K_w}{[OH^−]}  =  \frac{1.0×10^{−14}}{3.0×10^{−2}}  =  3.3×10^{−13}  M
          pH  =  −\log  (3.3×10^{−13})  =  12.48

Check The value of x seems reasonable: \sqrt{(∼6×10^{−4})(1.5)}  =  \sqrt{9×10^{−4}}  =  3×10^{−2}. Because (CH_3)_2NH is a weak base, the pH should be several pH units above 7.

Comment Alternatively, we can find the pOH first and then the pH:
          pOH  =  −\log  [OH^−]  =  −\log  (3.0×10^{−2})  =  1.52
          pH  =  14  −  pOH  =  14  −  1.52  =  12.48