When 200. mL of 2.0 \(M \ce{NaOH(aq)}\) is added to 500. mL of 1.0 \(M \ce{HCl(aq)}\), the pH of the resulting mixture is closest to

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The acid and base will react completely:

$$ \ce{NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)}$$ $$ \ce{OH-(aq) + H+(aq) -> H2O(l)} \tag*{net ionic equation} $$

Any excess hydroxide or hydronium ions will determine the pH.

$$ 0.5-0.4 = 0.1 \text{ mol excess } \ce{H+} $$

0.4 mol of each ion will react completely, and we are left with excess \(\ce{H+}\) ions. Although we are not allowed a calculator, we can use properties of logs to get the exact value:

$$ \text{pH} = -\log[\ce{H+}] $$ $$ \text{pH} = -\log_{10}(0.1) $$ $$ \text{pH} = \log_{10}\left(\frac{1}{10}\right)^{-1} $$ $$ \text{pH} = \log_{10}10 $$ $$ \text{pH} = 1 $$

Even without any calculations, since there is an excess of a strong acid, we can infer that the pH will be very low.