Question 7.5: The pH of Strong Bases Calculate the pH of a 5.0 × 10^-2 M N...

The major species in this solution are

Na^{+},                OH^{-},                    and                H _{2}O
From NaOH

Although the autoionization of water also produces OH^{-} ions, the pH will be determined by the OH^{-} ions from the dissolved NaOH. Thus in the solution

[OH^{-}] = 5.0 ×  10^{-2}   M

The concentration of H^{+} can be calculated from K_{w} :

[H^{+}]   =\frac{ K_{w} }{[OH^{-}]}= \frac{1.0 ×  10^{-14} }{5.0 ×  10^{-2} }  = 2.0 ×  10^{-13}  M

pH = 12.70

Note that this solution is basic:

  [OH^{-}]   >  [H^{+}]          and   pH > 7

The added OH^{-} from the salt has shifted the water autoionization equilibrium

H _{2}O (l)  \xrightleftharpoons[]{}  H^{+} (aq) + OH^{-} (aq)

to the left, significantly lowering the   [H^{+}]     compared with that in pure water.