A) 1 mL of pH = 2 is diluted to 100 mL
B) 0.01 mol of \[NaOH\] is added to 100 mL of 0.01 M \[NaOH\] solution
C) 100 mL of \[{{H}_{2}}O\] is added to 900 mL of \[{{10}^{-6}}\] M \[HCl\]
D) 100 mL of pH = 2 solution is mixed with 100 mL of pH = 12
Correct Answer: D
Solution :
| [d] | ||||||||||||||||||||
| ||||||||||||||||||||
| [a] \[pH=2.[{{H}^{+}}]={{10}^{-2}}M\] | ||||||||||||||||||||
| \[{{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}\] |
| \[{{10}^{-2}}\times 1={{M}_{2}}\times 100\] |
| \[{{M}_{2}}={{10}^{-4}}\] |
| \[\therefore \] \[pH=4\] |
| [b] \[\left[ O{{H}^{-}} \right]={{10}^{-2}}\] |
| \[pOH=2\] |
| \[pH=12\] 100 of 0.01 M \[NaOH\] |
| contains 0.001 mol |
| \[NaOH\] added - 0.01 mol |
| Total moles = 0.011 mol |
| \[\therefore \] \[{{\left[ O{{H}^{-1}} \right]}_{final}}=0.011\times 10=0.11M\] |
| \[pOH=0.96\] |
| pH = 13.04 |
| [c] \[\left[ HCl \right]={{10}^{-6}}M\] |
| \[\left[ {{H}^{+}} \right]={{10}^{-6}}M;pH=6\] |
| After dilution \[\left[ HCl \right]={{10}^{-7}}M=\left[ {{H}^{+}} \right]\] |
| \[\left[ {{H}^{+}} \right]in\,{{H}_{2}}O={{10}^{-7}}M\] |
| Total \[\left[ {{H}^{+}} \right]={{10}^{-7}}+{{10}^{-7}}=2\times {{10}^{-7}}\] |
| \[\therefore \] \[pH=7-\log 2=6.7\] |
| [d] \[\,pH=2,\,\left[ {{H}^{+}} \right]={{10}^{-2}}M\] |
| \[pH=12,\,\left[ {{H}^{+}} \right]={{10}^{-12}}M\] |
| \[\therefore \] \[\,\left[ O{{H}^{-}} \right]={{10}^{-2}}M\] |
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