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JEE Advanced levelSingle correct optionDifficulty 3 of 3ie150

50 mL of 0.20 M weak acid HA ( $p K_{a} = 3.90$ ) is mixed with 30 mL of 0.10 M NaOH. The pH of the resulting solution is

A4.27
B3.90
C3.53
D3.60

Answer: C

HA = 10 mmol, NaOH = 3 mmol. After reaction HA = 7 mmol, A

^{-}

= 3 mmol. pH =

3.90 + lo g (3/7) = 3.90 - 0.368 = 3.53

.

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