A) 8
B) 5
C) 6.9
D) None
Correct Answer: C
Solution :
| [c] pH of HCl solution = 5 |
| \[{{[{{H}^{+}}]}_{\text{initial}}}={{10}^{-5}}\] |
| \[{{[{{H}^{+}}]}_{\text{initial}}}\times {{V}_{1}}={{[{{H}^{+}}]}_{\text{final}}}\times {{V}_{2}}\] |
| \[{{[{{H}^{+}}]}_{\text{final}}}\times {{10}^{-8}}\] |
| Let the \[{{H}^{+}}\] from water be x. |
| \[{{H}_{2}}O\rightleftharpoons \underset{x+{{10}^{-8}}}{\mathop{{{H}^{+}}}}\,+\underset{x}{\mathop{O{{H}^{-}}}}\,\] |
| \[x(x+{{10}^{-8}})={{10}^{-14}}\] |
| \[{{x}^{2}}+{{10}^{-8}}-{{10}^{-14}}=0\] |
| \[x=\frac{-{{10}^{-8}}\pm \sqrt{{{10}^{-16}}=4\times {{10}^{-14}}}}{2}\] |
| \[\Rightarrow \] \[x=\frac{-{{10}^{-8}}+{{10}^{-8}}\sqrt{401}}{2}=19\times {{10}^{-8}}=9.5\times {{10}^{-8}}\] |
| \[{{[{{H}^{+}}]}_{\text{Total}}}={{[{{H}^{+}}]}_{\text{HCl}}}+{{[{{H}^{+}}]}_{\text{Water}}}\] |
| \[={{10}^{-8}}+{{10}^{-8}}\times 9.5=10.5\times {{10}^{-8}}\] |
| \[pH=8-\log \,10.5\] |
| Slightly less then 7 Hence [c]. |
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