Question

pOH of \[\dfrac{1}{{200}}mol{m^{ - 3}}\] \[{H_2}S{O_4}\left( {aq.} \right)\] solution at \[{25^ \circ }C\]
(A) \[2\]
(B) \[5\]
(C) \[9\]
(D) \[12\]

Answer 1

Hint: According to Arrhenius theory, acids are defined as the compounds that release hydrogen ions when dissolved in water. The concentration of hydrogen ions in a solution determines the strength of the acid. A pH scale is used to measure the acidic or basic strength. It is calculated by, taking the negative logarithm of hydronium ion concentration,
Formula used: \[pH = - {\log _{10}}\left[ {{H_3}{O^ + }} \right]\]
\[pOH = 14 - pH\]

Complete answer:
The concentration of hydronium ion in dilute solution of strong acid is equal to the concentration of the acids. The dissociation constants for strong acids are high and they fully dissociate in aqueous solution.
\[{H_2}S{O_4}\left( {aq.} \right)\] gets completely dissociated and produces two hydronium ions \[{H_3}{O^ + }\] due to which the concentration of hydronium ion in solution becomes:
\[\left[ {{H_3}{O^ + }} \right] = 2\left[ {{H_2}S{O_4}_{\left( {aq} \right)}} \right]\]
The concentration of \[{H_2}S{O_4}\left( {aq.} \right)\] is given in the question as \[\dfrac{1}{{200}}mol{m^{ - 3}}\]
\[\left[ {{H_3}{O^ + }} \right] = 2\left[ {\dfrac{1}{{200 \times 1000}}} \right]mol{L^{ - 1}}\]
Substituting the value of the concentration of hydronium ion in the pH calculation formula.
\[pH = - {\log _{10}}\left[ {2\left[ {\dfrac{1}{{2 \times {{10}^5}}}} \right]} \right]\]
Simplify the above obtained equation by using the properties of logarithm.
\[pH = - {\log _{10}}\left[ {{{10}^{ - 5}}} \right]\]
\[pH = 5{\log _{10}}\left( {10} \right)\]
\[pH = 5\]
The pH of the aqueous solution of dilute sulphuric acid is 5.
It is required to find the pOH of the solution. We know that at \[{25^ \circ }C\] the relation between pH and pOH is : \[pOH = 14 - pH\]
Substituting the value of pH in the above equation to find pOH.
\[pOH = 14 - 5\]
\[pOH = 9\]
Thus, the pOH of \[\dfrac{1}{{200}}mol{m^{ - 3}}\] \[{H_2}S{O_4}\left( {aq.} \right)\] solution at \[{25^ \circ }C\] is 9.
Option (c) 9, is the correct choice.

Note:
Remember that whenever a strong is taken into consideration first check the number of hydrogen it can release. As the concentration of hydronium ions depends on the concentration of hydrogen released by acid. As in the above case, $H_2SO_4$ tendency to give 2 hydrogen ions, another acid like hydrochloric acid tends to give one hydrogen ion. Then in the case of hydrochloric acid, the concentration of hydrogen ion remains the same as the acid and in sulphuric acid, the concentration of hydrogen ion becomes twice the acid.