A) \[7.04\]
B) \[9.37\]
C) \[9.26\]
D) \[8.37\]
Correct Answer: D
Solution :
0.005 M calcium acetate \[(C{{H}_{3}}{{(COO)}_{2}}Ca\] \[\underset{0.005\,\,M}{\mathop{{{(C{{H}_{3}}COO)}_{2}}Ca}}\,\xrightarrow{{}}C{{a}^{2+}}+\underset{(2\times 0.005=0.01)}{\mathop{2C{{H}_{3}}CO{{O}^{-}}}}\,\] \[\therefore \] \[[C{{H}_{3}}CO{{O}^{-}}]=0.01\,M\] \[C{{H}_{3}}CO{{O}^{-}}{{H}_{2}}O\overset{{}}{leftrightarrows}C{{H}_{3}}COOH+\underset{\text{Alkaline}}{\mathop{O{{H}^{-}}}}\,\] \[pH=7+\frac{p{{K}_{a}}}{2}+\frac{\log C}{2}\] \[=7+2.37+\frac{\log 0.0.1}{2}\] \[=7+2.37-1\] \[=8.37\]
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