Question #635c0

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Question #635c0

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Answer 1 · 2015-05-20T12:20:57

pH =4.56
The partial neutralization of acetic acid by sodium hydroxide leaves 0.0015 moles of acid and produces 0.001 moles of acetate.
The final solution is a buffer of acetic acid and acetate.
Final concentration of acetic acid ( $A c O H$ $AcOH$ ) = $0.1 M \cdot \frac{15}{35}$ $0.1 M *15/35$
=0.0428 M
Final concentration of $(O H^{-})$ $(OH^-)$ = acetateion( $A c O^{-}$ $AcO^-$ )= $0.1 M \cdot \frac{10}{35}$ $0.1M*10/35$ =0.0286 M
From the Henderson-Hasselbach equation:
$p H = p K a + log (\frac{A c^{-}}{H A c})$ $pH=pKa+log ((Ac^-)/(HAc))$
pH= $- log 1.8 \cdot 10^{- 5}$ $-log1.8*10^-5$ + $log (\frac{0.0287}{0.0428})$ $log (0.0287/0.0428)$ = 4.74-0.17 = 4.56

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