$20.0 mL$ of $0.100 M$ HCl is titrated with $0.100 M$ NaOH. The pH after $30.0 mL$ of NaOH has been added is

Answer: B

Excess

OH X^{-}

= 3.0 - 2.0 = 1.0 mmol

50.0 mL

, so

[OH X^{-}] = 0.020 M

pOH = 1.70

pH = 12.30

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