A) \[8.2\times {{10}^{-6}}\]
B) \[6.4\times {{10}^{-6}}\]
C) \[5.3\times {{10}^{-5}}\]
D) \[2.4\times {{10}^{-6}}\]
Correct Answer: A
Solution :
For complete neutralisation, Total mill equivalent of acid = mill equivalent of base\[=26.6\times 0.1=2.66\] For partial neutralisation \[HA+BOH\xrightarrow[{}]{{}}BA+{{H}_{2}}O\] 2.66 1.2 0 0 before reaction 1.46 0 1.2 1.2 after reaction The resultant mixture has HA and BA and thus act as buffer. \[\therefore \] \[pH=-\log {{K}_{a}}+\log \frac{[Salt]}{[Acid]}\] \[5=-\log {{K}_{a}}+\log \frac{1.2}{1.46}=-\log {{K}_{a}}-0.86\] \[\therefore \] \[\log {{K}_{a}}=-5.86\] or \[{{K}_{a}}=\text{anti}\,\,\log \,\,(-5.86)=8.21\times {{10}^{-6}}\]You need to login to perform this action.
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