25.0 mL of 0.10 M HCl is mixed with 20.0 mL of 0.080 M NaOH. Use $lo g 2 = 0.301$ . The pH of the mixture is closest to

Answer: C

HCl

= 2.5

mmol and NaOH

= 1.6

mmol, so 0.9 mmol of

H X^{+}

remains in 45 mL:

[H X^{+}] = 0.020

= 0.020

M. pH

= 2 - lo g 2 = 1.70

. Dividing by 25 mL instead of 45 mL gives 1.44.

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