A) \[0.019%\]
B) \[1.9%\]
C) \[3.0%\]
D) \[4.74%\]
Correct Answer: B
Solution :
\[C{{H}_{3}}COOHC{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}\] Initial 1 0 0 At equili \[1-\alpha \] \[\alpha \] \[\alpha \] \[p{{K}_{a}}=-\log {{K}_{a}}=4.74\] \[\therefore \] \[{{K}_{a}}=\text{antilog}(4.74)1.82\times {{10}^{-5}}\] 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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