Question

If 20 mL of an acetic acid of pH 3 is diluted to 100 mL, the ${H^ + }$ion concentration will be:
A. $1 \times {10^{ - 3}}$
B. $2 \times {10^{ - 3}}$
C. $2 \times {10^{ - 4}}$
D. $0.02 \times {10^{ - 4}}$

Answer 1

Hint
pH is given in the question so we can calculate ${H^ + }$ion concentration directly and then applying the law of equivalent for acetic acid and water we easily get the normality which is our required result.
- Formula Used:
We will have to use a pH formula which is used to know about the acidity or basicity.
pH = $- \log ({H^ + })$
Law of equivalent: We know $Normality = \dfrac{{No. of \, gram \, equivalent}}{{Volume \, of \, solution}}$
$No.{\text{ }}of{\text{ }}gram{\text{ }}equivalent{\text{ }}of{\text{ }}acid = No.{\text{ }}of{\text{ }}gram{\text{ }}equivalent{\text{ }}of{\text{ }}base\\\$
$N1V1 = N2V2$

Complete step by step solution:
Given: Volume of acetic acid ($V1$) = 20, of acetic acid = 3 and Volume of $H2O$ = 100
pH = - \log ({H^ + }) \\\
$\Rightarrow \;$ 3= - \log 10({H^ + }) \\\
$\Rightarrow \;{H^ + } = 1 \times {10^{ - 3}}$N

This is the concentration of ion concentration of acetic acid. Now let’s apply the law of equivalent to get the normality of.
1 \times {10^{ - 3}} \times 20 = N2 \times 100 \\\
\Rightarrow N2 = 2 \times {10^{ - 4}} \\\
This is the required concentration of ${H^ + }$ion.
**So the correct option is C.

Additional Information:**
pH is a measure of how acidic/basic water is. The range goes from 0 to 14, with 7 being neutral.
Of less than 7 indicate acidity, whereas of greater than 7 indicate a base. pH is a measure of free hydrogen and hydroxyl ion in water.

Note:
Always use the law of equivalent concept in case of dilution problems. Remember the concentration terms and their formulas; they help you to tackle these types of problems. Base of log is always 10.