Question

A bottle, which contains 200ml of 0.1M $KOH$, absorbs $1$ millimole of $C{O_2}$ from the air. If the solution is then treated with standard acid using phenolphthalein indicator, the normality will be
A. $0.095N$
B. $0.1N$
C. $0.2N$
D. $0.036N$

Answer 1

Hint: First we should write what happens when $KOH$ reacts with carbon dioxide. From this, we will get the number of millimoles of $KOH$ which is reacting. We can then equate the milli equivalent of acid used to the sum of millimoles of salt produced and the millimoles of $KOH$ which remains unreacted. As we know, normality is the number of milliequivalents divided by the volume of the solution.

Formulas used: $m.e{q_{acid}} = m.e{q_{unreactedKOH}} + m.e{q_{{K_2}C{O_3}}}$
Where $m.eq$ denotes the milliequivalent and the subscript denotes the component
$N = \dfrac{{m.e{q_{acid}}}}{V}$
Where $N$ is the normality, $m.e{q_{acid}}$ is the milliequivalents of acid and $V$ is the volume of the solution.

Complete step by step answer:
First let us calculate the number of millimoles of $KOH$ given. It is given that the concentration is $0.1M$. What this means is that there are $0.1mol = 0.1 \times {10^3} = 100mmol$ of $KOH$ in $1L$ of the solution. Hence, if we take the number of moles in $200mL = 0.2L$ as $x$, we get:

Number of moles	Volume of solution
$100mmol$	$1L$
$x$	$0.2L$

Using the cross-multiplication rule, we have:
$100 \times 0.2 = 1 \times x$
$ \Rightarrow x = 20mmol$
Let us now see what happens when $KOH$ absorbs carbon dioxide:
$2KOH + C{O_2} \to {K_2}C{O_3} + {H_2}O$
Thus, each millimole of carbon dioxide will react with two millimoles of $KOH$, as we can see from the stoichiometric equation.
But we have found out that the number of millimoles of $KOH = 20$
Hence, the amount of unreacted $KOH$ will be the difference between the total amount of $KOH$ and the amount which reacted with carbon dioxide ($2mmol$). Therefore, we have:
Unreacted $KOH$ $ = 20 - 2 = 18mmol$
Also, from the reaction, we see that absorbing one millimole of carbon dioxide will lead to the production of one millimole of ${K_2}C{O_3}$.
Since ${K_2}C{O_3}$ forms a strong base in solution, the acid used in titration will react with the unreacted $KOH$ as well as the ${K_2}C{O_3}$ formed during the absorption of carbon dioxide.
Hence, the number of milliequivalents of the acid used will be the sum of the milliequivalents of the unreacted $KOH$ and the ${K_2}C{O_3}$ which was produced. Therefore, we have:
$m.e{q_{acid}} = m.e{q_{unreactedKOH}} + m.e{q_{{K_2}C{O_3}}}$
Where $m.eq$ denotes the milliequivalents and the subscript denotes the component
Substituting the values as $m.e{q_{unreactedKOH}} = 18$ and $m.e{q_{{K_2}C{O_3}}} = 1$, we get:
$m.e{q_{acid}} = 18 + 1 = 19$
As we know,
$N = \dfrac{{m.e{q_{acid}}}}{V}$
Where $N$ is the normality, $m.e{q_{acid}}$ is the milliequivalents of acid and $V$ is the volume of the solution. Substituting our value as $m.e{q_{acid}} = 19$ and $V = 200mL$, we get:
$N = \dfrac{{19}}{{200}} = 0.095N$
Hence, the normality of the acid used will be $0.095N$.

So, the correct answer is Option A .

Note: Normality of a solution is defined as the number of equivalents of the substance per litre of the solution. The number of equivalents is in turn the ratio between the given mass of the substance and its equivalent weight. Note that in titrations, the indicator is used to identify the end point, normally accompanied by a change in colour of the solution.