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 milliequivalent of acid = milliequivalent of base \[=26.6\times 0.1=2.66\] For partial neutralisation, \[\begin{align} & HA+BOH\xrightarrow{{}}BA+{{H}_{2}}O \\ & 2.66\,\,\,\,\,12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\text{before}\,\text{reaction} \\ & \text{1}\text{.46}\,\,\,\,\,\,\,\text{0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{1}\text{.2}\,\,\,\,\,\,\,\,\text{1}\text{.2}\,\,\text{after}\,\text{reaction} \\ \end{align}\] 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 \[{{K}_{a}}=\text{anti}\,\text{log(-586)}\,\text{=}\,\text{8}\text{.21}\times {{10}^{-6}}\]
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