Question

20g of solid NaOH is dissolved in 800ml water. The solution is treated with $250ml\;0.8M\;HCl$ solution. The resulting solution is diluted to $2$ litre. $100ml$ of this diluted solution requires $V\;ml\;0.25M\;{H_2}S{O_4}$ for complete neutralisation. The value of $V$ is:
A) $600$
B) $30$
C) $60$
D) $100$

Answer 1

As we know that the normality of a solution is a measure of the concentration which is equal to the ratio of the gram equivalents of mass per litres of solution. We also know that gram equivalent mass is the reactive ability of a compound. For complete neutralisation we can compare the normalities of acid and base.

Formula used: $molarity = \dfrac{{mass\;of\;solute \times 1000}}{{molecular\;mass \times vol.\;of\;solvent(ml)}}$ and ${N_1}{V_1} = {N_2}{V_2}$.

Complete step by step solution:
As we are given that $20g$ sodium hydroxide is dissolved in $800ml$ water. So, let us first calculate its molarity which is the ratio of moles of solute to the volume of solvent. We will get:
$molarity = \dfrac{{mass\;of\;solute \times 1000}}{{molecular\;mass \times vol.\;of\;solvent(ml)}}$
${M_{NaOH}} = \dfrac{{20 \times 1000}}{{40 \times 800}}$
${M_{NaOH}} = 0.625M$

Now we will calculate the gram equivalents of sodium hydroxide, as $800ml$ of $0.625M\;NaOH$ contains sodium hydroxide:
$= \dfrac{{0.625 \times 800}}{{1000}} \\\ = 0.5\;g\;equivalent. \\\$
Now, we are given that this solution is treated with $250ml\;0.8M\;HCl$, so it contains hydrochloric acid:
$= \dfrac{{0.8 \times 250}}{{1000}} \\\ = 0.2\;g\;equivalent. \\\$
So, $0.2\,g\;equivalent.$ of $HCl$ would neutralize $0.2\,g\;equivalent.$ of $NaOH$ thus the limiting reagent is hydrochloric acid and the sodium hydroxide left in the solution is $0.5 - 0.2 = 0.3\;g\;equivalent.$ and the total volume of the solution would be: $800ml + 250ml = 1050ml$

Now we can calculate the normality of sodium hydroxide as we have the weight and the volume so it will be $= \dfrac{{0.3}}{{1.05}} = 0.286N$.
Now using this normality, we can dilute this solution in given $2000ml$ volume using
$\Rightarrow {N_1}{V_1} = {N_2}{V_2}$
$\Rightarrow 0.286 \times 1050 = {N_2} \times 2000$
$\Rightarrow {N_2} = 0.15N$
So, now we have the resulting solution and using its $100ml$ volume we can neutralise the sulphuric acid using the similar formula of normality where on the left hand side we have the normality and volume of the resulting solution of sodium hydroxide and on the right hand side we have sulphuric acid dimensions.
$\Rightarrow {N_1}{V_1} = {N_2}{V_2}$
$\Rightarrow 0.15 \times 100 = 2 \times 0.25 \times V$
As we know that the normality of sulphuric acid has a n-factor of $2$.
$V = 30ml$

Therefore, from the above explanation we can say that the correct answer is (B).

Note: Always remember that an acid and a base reacts with each other in equal proportions to neutralise the solution. These acids and bases completely dissociate into their ions whenever they are dissolved in water. Acid and base after reaction gives water and a salt in solution.