| \[HCO_{3}^{-}\] and \[0.2\,M\,CO_{3}^{2-}\] |
| \[[{{K}_{1}}({{H}_{2}}CO_{3}^{-})=4.2\times {{10}^{-7}}\times 10\]and |
| \[{{K}_{2}}(HCO_{3}^{-})=4.8\times {{10}^{-11}}].\] |
A) 3.18
B) 10.62
C) 6.62
D) 9.31
Correct Answer: B
Solution :
[b] \[HCO_{3}^{-}\to {{H}^{+}}+CO_{3}^{2-}\] \[{{K}_{2}}=\frac{[{{H}^{+}}][CO_{3}^{2-}]}{[HCO_{3}^{-}]}=4.8\times {{10}^{-11}}\] \[=4.8\times {{10}^{-11}}(0.1/0.2)\] \[pH=-\log [{{H}^{+}}]\] \[=-log\left( 4.8\times {{10}^{-11}}\times 0.5 \right)=10.62\]
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