A) \[3.2\times {{10}^{-3}}\]
B) \[7.8\times {{10}^{-6}}\]
C) \[7.8\times {{10}^{-28}}\]
D) \[3.2\times {{10}^{-4}}\]
Correct Answer: B
Solution :
| [b] |
| \[pH=12.4\] \[\Rightarrow \] \[-\log ({{H}^{+}})=12.4\,\Rightarrow \] |
| \[\log [{{H}^{+}}]=13.6\] |
| \[Ca{{(\overset{+}{\mathop{OH}}\,)}_{2}}\rightleftharpoons 2OH+C{{a}^{2+}}\] |
| \[[{{H}^{+}}]=4\times {{10}^{-13}}\] |
| \[[O{{H}^{-}}]=\frac{{{10}^{-14}}}{4\times 10}=\frac{1}{3}\times {{10}^{-1}}=2.5\times {{10}^{-2}}\] |
| \[Ksp=[C{{a}^{2+}}][O{{H}^{-}}]\] \[[C{{a}^{2+}}]=\frac{1}{2}[O{{H}^{-}}]\] |
| \[\left( \frac{1}{8}\times {{10}^{-1}} \right){{\left( \frac{1}{4}\times {{10}^{-1}} \right)}^{2}}=\frac{1}{8}\times \frac{1}{16}\times {{10}^{-3}}\] |
| \[7.8\times {{10}^{-6}}\] |
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