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![\text{NaOH}(aq) \longrightarrow \text{Na}^{+} (aq)+ \text{OH}^{-}(aq)\\\\k_w = [\text{H}_3 \text{O}^{+}][\text{OH}^{-}]\\\\\implies 10^{-14} = [\text{H}_3 \text{O}^{+}] \cdot 4\\\\\implies [\text{H}_3 \text{O}^{+}] = \dfrac{10^{-14}}4 = 2.5 \times 10^{-15}~ M\\\\\text{pH} = -\log[\text{H}_3 \text{O}^{+}] = -log(2.5 \times 10^{-15})=14.603](/tpl/images/2579/0555/5f2ae.png)
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Calculate the pH of the 4M NaOH...
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