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Question: To a 10 mL of \[{10^{ - 3}}{\text{N }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\] sol...

To a 10 mL of 10−3N H2SO4{10^{ - 3}}{\text{N }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} solution, water has been added to make the total volume of one litre. Its pOH would be:
A.3
B.12
C.9
D.5

Explanation

Solution

The pOH of the solution of the “power of hydroxyl ions” is the solution just like pH is the “power of hydronium ions” of the solution. We shall first calculate the pH of the given solution, then using the formula, we shall calculate the pOH of the solution.
Formula used:
pOH = −log[OH] - {\text{pOH = }} - {\text{log}}{\left[ {{\text{OH}}} \right]^{\text{ - }}} and also, pH+pOH=14{\text{pH}} + {\text{pOH}} = 14

Complete step by step answer:
The pH of a 10 mL 10−3N H2SO4{10^{ - 3}}{\text{N }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}solution will be:
pH=−log10[H + ]{\text{pH}} = - {\text{lo}}{{\text{g}}_{{\text{10}}}}\left[ {{{\text{H}}^{\text{ + }}}} \right], now the H + {{\text{H}}^{\text{ + }}} ion concentration in the solution will be:
10×10−31000=10−5\dfrac{{10 \times {{10}^{ - 3}}}}{{1000}} = {10^{ - 5}} mole/litre.
Therefore, the pH of the solution =−log10[10−5] = - {\text{lo}}{{\text{g}}_{{\text{10}}}}\left[ {{{10}^{ - 5}}} \right] = −(−5)log1010=5 - \left( { - 5} \right){\text{lo}}{{\text{g}}_{{\text{10}}}}10 = 5
Now according to the above formula,
pH+pOH=14{\text{pH}} + {\text{pOH}} = 14
Therefore, pOH=14−5=9{\text{pOH}} = 14 - 5 = 9

Hence, the correct answer to the above question is option C.

Note:
The pOH scale is similar to the pH scale where, a pOH value of 7 indicates a neutral solution, a basic solution has a pOH less than 7 while an acidic solution has a pOH greater than 7.
The values of some commonly available substances are: Battery acid (pOH = 14), stomach acid (pOH = 13), lemon juice (pOH = 11), soda water (pOH = 10), Household ammonia (pOH = 2).
The formula, pH+pOH=14{\text{pH}} + {\text{pOH}} = 14has been deduced from the ionic product of water that comes from the self-ionization of water.
2H2O⇄H3O +  + OH - 2{{\text{H}}_{\text{2}}}{\text{O}} \rightleftarrows {{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{ + O}}{{\text{H}}^{\text{ - }}}
From the above equation, kw = [H3O + ][OH - ]{{\text{k}}_{\text{w}}}{\text{ = }}\left[ {{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}} \right]\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]
where kw{{\text{k}}_{\text{w}}}is the ionic product of water whose value at 250C{25^0}{\text{C}}is equal to1×10−141 \times {10^{ - 14}}.
Now, taking log on both sides of the above equation:
−log⁡10[10−14]=−log⁡10[H3O+]−log⁡10[OH+]- {\log _{10}}\left[ {{{10}^{ - 14}}} \right] = - {\log _{10}}\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] - {\log _{10}}\left[ {{\text{O}}{{\text{H}}^ + }} \right]
⇒pH+pOH=14\Rightarrow {\text{pH}} + {\text{pOH}} = 14