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Question: An aqueous solution contains \[{10^{ - 4}}[{H^ + }]\]. If it is diluted by mixing equal volumes of w...

An aqueous solution contains 10−4[H+]{10^{ - 4}}[{H^ + }]. If it is diluted by mixing equal volumes of water then the concentration of OH−O{H^ - } in mol dm−3d{m^{ - 3}} will be:
A. 0.5×10−100.5 \times {10^{ - 10}}
B. 2×10−102 \times {10^{ - 10}}
C. 10−6{10^{ - 6}}
D. 10−8{10^{ - 8}}

Explanation

Solution

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.
Formula used: M1V1=M2V2{M_1}{V_1} = {M_2}{V_2} and pH=−log⁡[H+]pH = - \log [{H^ + }]

Complete step by step answer:
We are given the concentration of aqueous solution having hydrogen ions =10−4M = {10^{ - 4}}M
Let us suppose V 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 2V. the new concentration can now be calculated by using the formula of dilution. It is
M1V1=M2V2{M_1}{V_1} = {M_2}{V_2}
We know that M1=10−4M{M_1} = {10^{ - 4}}M , V1=V{V_1} = V , V2=2V{V_2} = 2V , so M2{M_2} can be calculated by putting these values in above formula.

10−4×V=M2×2V ∴M2=5×10−5M  {10^{ - 4}} \times V = {M_2} \times 2V \\\ \therefore {M_2} = 5 \times {10^{ - 5}}M \\\

This is the new concentration of hydrogen ion [H+][{H^ + }].
Now, we know the formula of calculating pH of solution. pH is the negative logarithm of the hydrogen ion concentration or the amount of hydrogen ions present in a solution. we can write it as-
pH=−log⁡[H+]pH = - \log [{H^ + }]
We can substitute the new concentration of hydrogen ions in it and get,

pH=−log⁡[5×10−5] ∴pH=4.3  pH = - \log [5 \times {10^{ - 5}}] \\\ \therefore pH = 4.3 \\\

Water undergoes a measurable equilibrium at 298K, so the sum of pH and pOH is always 14. This is because the product of proton and hydroxide concentration must always be equal to the equilibrium constant for water ionisation. We can depict it as-
pH+pOH=14pH + pOH = 14
So, pOH=14−pHpOH = 14 - pH
We calculated the pH which is equal to 4.3, therefore pOH will be

pOH=14−4.3 ∴pOH=9.69  pOH = 14-4.3 \\\ \therefore pOH = 9.69 \\\

Since pOH is the negative algorithm of hydroxide ion concentration, we can further write it as
pOH=−log⁡[OH+]pOH = - \log [O{H^ + }]
Thus, we can say that

\-log⁡[OH]−=9.69 [OH]−=10−9.69=2×10−10  \- \log {[OH]^ - } = 9.69 \\\ {[OH]^ - } = {10^{ - 9.69}} = 2 \times {10^{ - 10}} \\\

Therefore, the final concentration of [OH]−=2×[10−10{[OH]^ - } = 2 \times [{10^{ - 10}} mol dm−3d{m^{ - 3}}.

Hence, the correct option is (B).

Note:
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.