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Answer 1 · 2015-12-10T07:25:41

$[H^{+}] = 7.58 \times 10^{- 6} M$ $[H^+]=7.58xx10^(-6)M$

We console this question using Henderson-Hasselbalch equation:

$p H = p K_{a} + log (\frac{[C_{2} H_{3} O_{2}^{-}]}{[H C_{2} H_{3} O_{2}]})$ $pH=pK_a+log(([C_2H_3O_2^-])/([HC_2H_3O_2]))$

$p K_{a} = - {log K}_{a} = - log (1.8 \times 10^{- 5}) = 4.74$ $pK_a=-logK_a=-log(1.8xx10^(-5))=4.74$

$\Rightarrow p H = 4.74 + log (\frac{0.48}{0.2}) = 5.12$ $=>pH=4.74+log((0.48)/(0.2))=5.12$

Now we can calculate the concentration of $H^{+}$ $H^+$ by:

$[H^{+}] = 10^{- p H} = 10^{- 5.12} = 7.58 \times 10^{- 6} M$ $[H^+]=10^(-pH)=10^(-5.12)=7.58xx10^(-6)M$

Here is a video that explains Henderson-Hasselbalch equation:
Acid - Base Equilibria | Buffer Solution.