When 200. mL of 2.0
M N a O H ( a q ) M \ce{NaOH(aq)} M NaOH ( aq ) is added to 500. mL of 1.0
M H C l ( a q ) M \ce{HCl(aq)} M HCl ( aq ) , the pH of the resulting mixture is closest to
The acid and base will react completely:
N a O H ( a q ) + H C l ( a q ) → N a C l ( a q ) + H X 2 O ( l ) \ce{NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)} NaOH ( aq ) + HCl ( aq ) NaCl ( aq ) + H X 2 O ( l ) O H X − ( a q ) + H X + ( a q ) → H X 2 O ( l ) net ionic equation \ce{OH-(aq) + H+(aq) -> H2O(l)} \tag*{net ionic equation} OH X − ( aq ) + H X + ( aq ) H X 2 O ( l ) net ionic equation Any excess hydroxide or hydronium ions will determine the pH.
Hydroxide
2.0 M × 200 m L = 0.4 m o l O H X − 2.0 M \times \pu{200 mL} = \pu{0.4 mol } \ce{OH-} 2.0 M × 200 mL = 0.4 mol OH X − Hydronium
1.0 M × 500 m L = 0.5 m o l H X + 1.0 M \times \pu{500 mL} = \pu{0.5 mol } \ce{H+} 1.0 M × 500 mL = 0.5 mol H X +
0.5 − 0.4 = 0.1 mol excess H X + 0.5-0.4 = 0.1 \text{ mol excess } \ce{H+} 0.5 − 0.4 = 0.1 mol excess H X + 0.4 mol of each ion will react completely, and we are left with excess H X + \ce{H+} H X +
pH = − log [ H X + ] \text{pH} = -\log[\ce{H+}] pH = − log [ H X + ] pH = − log 10 ( 0.1 ) \text{pH} = -\log_{10}(0.1) pH = − log 10 ( 0.1 ) pH = log 10 ( 1 10 ) − 1 \text{pH} = \log_{10}\left(\frac{1}{10}\right)^{-1} pH = log 10 ( 10 1 ) − 1 pH = log 10 10 \text{pH} = \log_{10}10 pH = log 10 10 pH = 1 \text{pH} = 1 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.