Question

Equal volumes of two solutions having pH 2 and 6 are mixed. The pH of the resulting solution is approximately.
(a) 4
(b) 4.3
(c) 2.3
(d) 5

Answer 1

Hint: The pH is defined as the negative logarithm of hydrogen ion concentration present in the solution. To find out the resulting pH of the solution first the number of moles present in each solution is calculated by using the formula of molarity.

Complete answer:

Given:
Solution 1 has pH 2.
Solution 2 has pH 6.
The pH is the molar concentration of the hydrogen ion present in the solution.
The molar concentration of pH 2 is ${10}^{-2}M$
The molar concentration of pH 6 is ${10}^{-6}M$
The formula for calculating the moles is shown below.
$M={\displaystyle \frac{n}{V}}$
Where,
M is the molarity
n is the number of moles
V is the volume.
Let assume the volume of both the solutions be V.
To calculate the number of moles of solution 1, substitute the values in the above equation.
$\Rightarrow {10}^{-2}={\displaystyle \frac{n_1}{V}}$
$\Rightarrow n_1={10}^{-2}\times V$
To calculate the number of moles of solution 2, substitute the values in the above equation.
$\Rightarrow {10}^{-6}={\displaystyle \frac{n_2}{V}}$
$\Rightarrow n_2={10}^{-6}\times V$
The net number of moles of solution 1 and solution 2 after mixing is
$\Rightarrow {10}^{-2}\times V+{10}^{-6}\times V$
The net molar concentration of $H^{+}$ ion is given as shown below.
$[H^{+}]={\displaystyle \frac{{10}^{-2}\times V+{10}^{-6}\times V}{2V}}$
$\Rightarrow [H^{+}]={\displaystyle \frac{V[{10}^{-2}+{10}^{-6}]}{2V}}$
And hence on doing the simplification, we have
$\Rightarrow [H^{+}]={\displaystyle \frac{[{10}^{-2}+{10}^{-6}]}{2}}$
$\Rightarrow [H^{+}]=5.0005\times {10}^{-3}$
Then the pH is calculated as shown below.
$\Rightarrow pH=-\log [5.0005\times {10}^{-3}]$
$\Rightarrow pH=2.3$
Thus, the pH of the resulting solution is approximately 2.3.

Therefore, the correct option is c.

Note: Make sure that we have to find out the hydrogen ion concentration of both the solution after mixing the solution of different pH. As, the resulting pH is 2.3 the solution is acidic in nature.