A) 0.019%
B) 1.9%
C) 3.0%
D) 4.74%
Correct Answer: B
Solution :
\[\underset{\begin{smallmatrix} lnitial\,1 \\ \text{At}\,\text{equili}\,1-\alpha \end{smallmatrix}}{\mathop{C{{H}_{3}}COOH}}\,\underset{\alpha }{\mathop{\underset{0}{\mathop{C{{H}_{3}}CO{{O}^{-}}}}\,}}\,+\underset{\alpha }{\mathop{\underset{0}{\mathop{{{H}^{+}}}}\,}}\,\] \[p{{K}_{a}}=-\log \,{{K}_{a}}=4.74\] \[\therefore \] \[{{K}_{a}}=\text{antilog(4}\text{.74)1}\text{.82}\times {{10}^{-5}}\] From \[{{K}_{a}}=\frac{C{{\alpha }^{2}}}{(1-\alpha )}=C{{\alpha }^{2}}\] \[(1-\alpha \approx 1)\] \[\alpha =\sqrt{\frac{{{K}_{a}}}{C}}=\sqrt{\frac{1.82\times {{10}^{-5}}}{0.05}}\] \[=0.019.\]or \[1.9\,%\]
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