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Question: The pH of the solution obtained by mixing 100 ml of a solution of pH = 3 with 400 mL of a solution o...

The pH of the solution obtained by mixing 100 ml of a solution of pH = 3 with 400 mL of a solution of pH = 4 is
A. 3−log⁡2.83 - \log 2.8
B. 7−log⁡2.87 - \log 2.8
C. 4−log⁡2.84 - \log 2.8
D. 5−log⁡2.85 - \log 2.8

Explanation

Solution

We have to calculate the concentration of solution at pH 3 and pH 4. Using the concentration and volume of both solutions, we have to calculate the concentration of the resulting solution. From the obtained concentration, we have to calculate the pH of the solution using the formula,
pH=−log⁡[H+]pH = - \log \left[ {{H^ + }} \right]
Here [H+]\left[ {{H^ + }} \right] is the concentration of hydrogen ions.

Complete step by step answer:
Given data contains,
For solution 1, the pH is 3.
For solution 1, the volume (V1)\left( {{V_1}} \right) is 100ml.
For solution 2, the pH is 4.
For solution 2, the volume (V2)\left( {{V_2}} \right) is 400ml.
We know that concentration of solution would be obtained by taking the antilog of pH.
We can write the formula to calculate the concentration of solution as,
[H+]=10−pH\left[ {{H^ + }} \right] = {10^{ - pH}}
Let us now substitute the pH of the solution 1 in the expression.
[H+]=10−3\left[ {{H^ + }} \right] = {10^{ - 3}}
So, the concentration of solution 1 is 10−3M{10^{ - 3}}M.
Let us now substitute the pH of the solution 2 in the expression.
[H+]=10−4\left[ {{H^ + }} \right] = {10^{ - 4}}
So, the concentration of solution 2 is 10−4M{10^{ - 4}}M.
We can calculate the concentration of resultant solution using the formula,
M=M1V1+M2V2V1+V2M = \dfrac{{{M_1}{V_1} + {M_2}{V_2}}}{{{V_1} + {V_2}}}
Let us now substitute the values of concentration and volumes of both solutions in the expression.
M=M1V1+M2V2V1+V2M = \dfrac{{{M_1}{V_1} + {M_2}{V_2}}}{{{V_1} + {V_2}}}
M=10−3×100+10−4×400100+400M = \dfrac{{{{10}^{ - 3}} \times 100 + {{10}^{ - 4}} \times 400}}{{100 + 400}}
M=0.14500M = \dfrac{{0.14}}{{500}}
M=2.8×10−4MM = 2.8 \times {10^{ - 4}}M
So, we have calculated the molarity of the resultant solution as 2.8×10−4M2.8 \times {10^{ - 4}}M. From this, let us now calculate the pH of the solution.
We have to calculate the pH of the solution using the formula,
pH=−log⁡[H+]pH = - \log \left[ {{H^ + }} \right]
Let us now substitute the value of concentration of solution.
pH=−log⁡[H+]pH = - \log \left[ {{H^ + }} \right]
pH=−log⁡[2.8×10−4]pH = - \log \left[ {2.8 \times {{10}^{ - 4}}} \right]
pH=−log⁡(2.8)−log⁡(10−4)pH = - \log \left( {2.8} \right) - \log \left( {{{10}^{ - 4}}} \right)
pH=4−log⁡2.8pH = 4 - \log 2.8
So, the pH is 4−log⁡2.84 - \log 2.8.

Therefore, the option C is correct.

Note: Generally solutions that are acidic are known to contain lesser values of pH when compared to solutions that are alkaline (or) basic. The pH is a dimensionless quantity and pH scale is logarithmic. It is important to measure pH in chemistry, medicine, treatment of water and several other applications.