Question

A bottle which contains $200{\text{ ml}}{\text{. of 0}}{\text{.1 M NaOH}}$absorbs $1.0{\text{ mmol of C}}{{\text{O}}_2}$ from the air. If the solution is then titrated with standard acid using phenolphthalein indicator, what normality of the acid will be found ?
(A) $0.190\;N$
(B) $0.380\;N$
(C) $0.095\;N$
(D) $0.0475\;N$

Answer 1

Hint: Normality is the number of grams or mole equivalents of solutes present per liter of solution.
Normality $ = \dfrac{{{\text{Gram equivalents of solute}}}}{{{\text{Volume of solution in litre}}}}$

Complete step by step answer:
Given , $200{\text{ ml of 0}}{\text{.1 M NaOH}}$ absorbs $1.0\;{\text{mmol of C}}{{\text{O}}_2}$.
Therefore, moles of $NaoH$present in the solution $ \Rightarrow $
Molarity of $\operatorname{NaoH} = 0.1M$
Volume of $\operatorname{NaoH} = 200ml$
          $ = \dfrac{{200}}{{1000}}L = 0.200L$
Number of moles $(n) = {\text{Molarity }} \times {\text{ Volume}}$
          $ = 0.1 \times 0.200$
          $ = 0.02{\text{ mol}}{\text{.}}$
The reaction will proceed like :-
$2{\text{NaoH + C}}{{\text{O}}_2} \to N{a_2}C{O_3} + {H_2}O.$
According to the equation, $1.0{\text{ mmol or 0}}{\text{.001 mol of C}}{{\text{O}}_2}$ will react with 2 molecules of ${\text{NaoH}}$.
Therefore $0.001{\text{ mol}}$ carbon dioxide reacts with $0.001 \times 2 = 0.002{\text{ mol of NaOH}}$ to from $0.001{\text{ mol}}$of sodium carbonate
After this reaction, number of moles of ${\text{NaoH left = 0}}{\text{.02 - 0}}{\text{.002}}$
                         $ = 0.018{\text{ mol}}{\text{.}}$
These $0.018{\text{ mol of NaoH}}$will react with $\operatorname{HCl} $ in titration using phenolphthalein indicator .
As phenolphthalein indicates the $50\% $ completion of reaction, gram equivalents of ${\text{HCl = }}\dfrac{{0.002}}{2}$ equivalents of ${\text{N}}{{\text{a}}_2}C{O_3}$
$ = 0.001$ equivalents of ${\text{HCl}}$.
Thus, Normality of the solution will be calculated as :-
$\operatorname{Normality} = \dfrac{{gram\;equivalents\;of\;NaoH\; + HCl}}{{Volume\;of\;solution\;in\;litre}}$
     $ = \dfrac{{0.018 + 0.001}}{{0.2}}$
     $ = 0.095{\text{ mol/L}}{\text{.}}$
Or $ = 0.095{\text{ N}}{\text{.}}$

Hence, the correct option is $\left( C \right)0.095N$.

Note:
For calculating the normality, we must first find out the gram equivalents of each solute from their given concentration. Gram equivalents are calculated by dividing the mass of solute by the number of equivalents per mole of solute.
Also, the units should be converted properly, example :-
Converted from $\operatorname{mL} of L \left( {litre} \right)$.