Question

The volume of $0.2N$ ${H_2}S{O_4}$ mixed with $0.5N$$KOH$ (remaining volume) to prepare $150c{m^3}$ of $0.2N$ $KOH$solution is :
A. $50c{m^3}$
B. $60c{m^3}$
C. $70c{m^3}$
D. $80c{m^3}$

Answer 1

Hint: We have studied in chemistry practicals that In acid base titration addition of a solution of known concentration is added to the another solution of unknown concentration until the reaction approaches its neutralization. Normality can be defined as the gram equivalent weight of solute per litre of solution. It is a unit of concentration of a chemical solution.



Complete step by step solution: We can solve this problem with the help of concept that when an acid of volume ${V_1}$ $c{m^3}$having normality ${N_1}$ is mixed with base of volume ${V_2}$ $c{m^3}$ and normality ${N_2}$ then the normality of the resultant solution will be \[N = \dfrac{{{N_1}{V_1} - {N_2}{V_2}}}{V}\] where \[{N_1}{V_1} > {N_2}{V_2}\].
Here in the given problem we have normality of sulphuric acid ${N_1}$ $ = 0.2N$ and normality of base i.e. potassium hydroxide ${N_2}$$ = 0.5N$.
 The number of equivalents of the final solution is $NV = 0.2 \times 150 = 30$ .
Let assume that the volume of sulphuric acid is $xc{m^3}$ then the volume of potassium hydroxide will be $(150 - x)$ $c{m^3}$. As we know that the resultant solution is basic so the value of ${N_2}{V_2}$ is greater than ${N_1}{V_1}$ so we will subtract ${N_1}{V_1}$ from ${N_2}{V_2}$.
We know that for the neutralisation of acid and bases we have; $NV = {N_1}{V_1} - {N_2}{V_2}$ $\begin{gathered}
  0.5 \times (150 - x) - 0.2 \times x = 0.2 \times 150 \\
  75 - 0.7x = 30 \\
  x = \dfrac{{45}}{{0.7}} = 64.2 \simeq 60 \\
\end{gathered} $
Hence volume of sulphuric acid is $60c{m^3}$. So the volume of $0.2N$ ${H_2}S{O_4}$ mixed with $0.5N$$KOH$ (remaining volume) to prepare $150c{m^3}$ of $0.2N$ $KOH$solution is $60c{m^3}$.

Here option B is the correct answer to this problem.



Note: We have approached this problem with the help of formula of acid base titration. The volume of the solution and the normality of acid, base and resultant solution is given then the calculation of the volume of acid becomes more easy.