A) 1.94, 1.85
B) 1.94, 7.94
C) 7.94, 1.85
D) 1.94, 8.25
Correct Answer: C
Solution :
\[{{\operatorname{CH}}_{2}}ClCOOH+{{H}_{2}}O\,\,\,\,C{{H}_{2}}ClCO{{O}^{-}}\,\,+\,\,{{H}_{3}}{{O}^{+}}\]\[{{\operatorname{K}}_{a}}=\,\,1.35\times 1{{0}^{-}}^{3}\] \[\operatorname{p}Ka = -log\,{{K}_{a}} = -\,log \left[ 1.35 \times 1{{0}^{-\,3}} \right] = 2.87\] By Ostwal?d dilution law, \[[{{H}_{3}}{{O}^{+}}]\,\,=\,\,\sqrt{{{K}_{a}}\cdot C}\,=\,\,\sqrt{1.35\times {{10}^{3}}\times 0.1}\,\,=\,\,1.16\,\times \,{{10}^{-\,2}}\,M\]\[pH\,\,of\,\,acid=-\log [{{H}_{3}}{{O}^{+}}]\,\,=\,\,1.94\] 0.1 \[M C{{H}_{2}}ClCOONa\] is basic due to hydrolysis. \[{{\operatorname{CH}}_{2}}ClCOONa+{{H}_{2}}O\,\,=\,\,C{{H}_{2}}ClCOOH+NaOH\]for a salt of strong bse + weak acid \[pH=7\times \frac{p{{K}_{a}}+\log C}{2}=7+\frac{2.87+\log \,\,0.1}{2}\]\[\operatorname{pH} = 7.94\]
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