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Question: The \( {{H}^{+}} \) ion concentration of a solution is \( 4\times {{10}^{-5}}M \) Then the \( O{{H}^...

The H+{{H}^{+}} ion concentration of a solution is 4×10−5M4\times {{10}^{-5}}M Then the OH−O{{H}^{-}} ion concentration of the same solution is:
(A) 4×10−5M4\times {{10}^{-5}}M
(B) 2.5×10−9M2.5\times {{10}^{-9}}M
(C) 1.0×10−7M1.0\times {{10}^{-7}}M
(D) 2.5×10−10M2.5\times {{10}^{-10}}M

Explanation

Solution

We know that we can solve for the concentration or volume of the concentrated or dilute solution by using the formula of dilution. Once we know the new concentration of hydrogen ions, we can calculate pH of the solution and thereby hydroxide ion concentration.

Complete step by step solution:
We are given the concentration of aqueous solution having hydrogen ions. Let us suppose VV is the volume of water present in it already. On further dilution, we add equal volume of water to it, so the final volume of solution now becomes double. The new concentration can now be calculated by using the formula of dilution. Water undergoes a measurable equilibrium at 298K298K , so the sum of pHpH and pOHpOH is always 14.14. This is because the product of proton and hydroxide concentration must always be equal to the equilibrium constant for water ionization. We can depict it as;
For any solution,
[H+][OH−]=10−14\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]={{10}^{-14}}
Substituting the values provided;
⇒[4×10−5][OH−]=10−14\Rightarrow \left[ 4\times {{10}^{-5}} \right]\left[ O{{H}^{-}} \right]={{10}^{-14}}
On further evaluating;
⇒[OH−]=4×10−5×10−14\Rightarrow \left[ O{{H}^{-}} \right]=4\times {{10}^{-5}}\times {{10}^{-14}}
Thus, [OH−]=2.5×10−10M\left[ O{{H}^{-}} \right]=2.5\times {{10}^{-10}}M
Therefore, the correct answer is option D.

Note:
Remember that on dilution, the amount of solute does not change, before and after dilution the number of moles remains the same. Only change in volume occurs on adding more solvent for dilution and the solute particles now have more space to move.