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Question: The \(pH\) of a solution is \(10\) and that of another is \(12\). When equal volumes of these two ar...

The pHpH of a solution is 1010 and that of another is 1212. When equal volumes of these two are mixed, the pHpH of the resulting solution is :
A.) 1010
B.) 1212
C.) 1111
D.) 10.301010.3010

Explanation

Solution

This question can be solved by the concept that when equal volumes of two solutions are mixed together then the concentration of hydrogen ion that is [H+][{H^ + }] of each solution will get half. Then the sum of these concentrations gives us total concentration of the resulting solution.

Complete step by step answer:
In this question, the pHpH value of the solution can be defined as the negative of the logarithm of the hydrogen ion. It is the measure of the acidity of the solution or the alkalinity of the solution. If the pHpH of the solution is less than seven then it is an acidic solution, if it is equal to seven then it is a neutral solution and if pHpH of the solution is more than seven then it is a basic solution. The pHpH of the solution can be written as:
pH=−log⁡[H+]pH = - \log [{H^ + }]
Or we can also write it as,
[H+]=10−pH[{H^ + }] = {10^{ - pH}}
Now, for the first solution we have given the pH=10pH = 10, therefore the value of concentration of hydrogen ion for solution −1 - 1 is [H+]=10−10[{H^ + }] = {10^{ - 10}}
Also, for the second solution we have given the pH=12pH = 12, therefore the value of concentration of hydrogen ion for solution −2 - 2 is [H+]=10−12[{H^ + }] = {10^{ - 12}}
As we know that when the equal volume of two solutions are mixed together then the concentration of [H+][{H^ + }] for each solution gets half. Therefore, the concentration of hydrogen ion for final solution is given as:
[H+]=[H+]12+[H+]22 [H+]=10−102+10−122 [H+]=10−10×(0.5+0.005) [H+]=0.505×10−10  [{H^ + }] = \dfrac{{{{[{H^ + }]}_1}}}{2} + \dfrac{{{{[{H^ + }]}_2}}}{2} \\\ [{H^ + }] = \dfrac{{{{10}^{ - 10}}}}{2} + \dfrac{{{{10}^{ - 12}}}}{2} \\\ [{H^ + }] = {10^{ - 10}} \times (0.5 + 0.005) \\\ [{H^ + }] = 0.505 \times {10^{ - 10}} \\\
Now, pHpH of the resulting solution is :
pH=−log⁡[0.505×10−10] pH=−log⁡[0.505]+log⁡[10−10] pH=−−0.30−10 pH=10.30  pH = - \log [0.505 \times {10^{ - 10}}] \\\ pH = - \\{ \log [0.505] + \log [{10^{ - 10}}]\\} \\\ pH = - \\{ - 0.30 - 10\\} \\\ pH = 10.30 \\\
As the required pHpH value of the resultant solution is approximately 10.3010.30.

Hence, option D.) is the correct answer.

Note:
Always remember that when equal volumes of two solutions are mixed together then the resultant is not the average of the pHpH values of both solutions but rather it is the average of the concentration of hydrogen ion that is given as [H+][{H^ + }].