Question

What is the volume of water that should be added to $ 150{\text{ }}ml $ of $ \dfrac{N}{2} $ oxalic acid to prepare a solution of $ \dfrac{N}{{10}} $ oxalic acid?
(a) $ 750c.c. $
(b) $ 400c.c. $
(c) $ 800c.c. $
(d) $ 600c.c. $

Answer 1

Hint :Here the oxalic acid is given in terms of normality. Where normality is strength multiplied by equivalent weight. The formula used here will be normality multiplied by the volume. This formula is only used for preparation of standard solutions during titration.

Complete Step By Step Answer:
We are given here with volume of oxalic acid $ 150{\text{ }}ml $ and concentration of oxalic acid is given in terms of normality $ \dfrac{N}{2} $ . We have to find out the total volume of water that will be required for the preparation of $ \dfrac{N}{{10}} $ oxalic acid. The formula that will be used here is normality multiplied by volume. The formula is
 $ \Rightarrow $ $ {N_1}{V_1}{\text{ }} = {\text{ }}{N_2}{V_2} $
This formula is only used in acid-base titration.
Where $ {N_1} $ is concentration of oxalic acid in terms of normality of first solution and $ {V_1} $ is the volume of $ \dfrac{N}{2} $ oxalic acid, $ {N_2} $ is the normality of second solution and $ {V_2} $ is the volume of the $ \dfrac{N}{{10}} $ oxalic acid.
Now let us calculate the volume of water required.
So $ \Rightarrow $ $ {N_1}{V_1}{\text{ }} = {\text{ }}{N_2}{V_2} $
 $ \Rightarrow $ $ {V_2} = {V_1} + x $
Given that $ {N_1} = \dfrac{N}{2} $ , $ {V_1} = 150ml $ , $ {N_2} = \dfrac{N}{{10}} $ , and $ {V_2} $ we will write as $ {V_1} + x $ where x is the volume of water added.
Thus, $ \Rightarrow $ $ {N_1}{V_1}{\text{ }} = {\text{ }}{N_2}{V_2} $
 $ {N_1}{V_1}{\text{ }} = {\text{ }}{N_2}\left( {{V_1} + x} \right) $
 $ \dfrac{N}{2} \times 150{\text{ }} = {\text{ }}\dfrac{N}{{10}}\left( {150 + x} \right) $
 $ \dfrac{1}{2} \times 150{\text{ }} = {\text{ }}\dfrac{1}{{10}}\left( {150 + x} \right) $
 $ 150{\text{ }} = {\text{ }}\dfrac{1}{5}\left( {150 + x} \right) $
 $ 5 \times 150{\text{ }} = {\text{ }}\left( {150 + x} \right) $
 $ x = 750 - 150 $
 $ \Rightarrow $ $ x = 600c.c. $
Hence the volume of water required for preparation of a solution of $ \dfrac{N}{{10}} $ oxalic acid is $ 600c.c. $
Therefore the correct option is $ D.\;\;\;\;\;600{\text{ }}c.c. $ .

Note :
Normality is described as the number of gram or mole equivalent of solute present in one liter of a solution. Equivalent weight means the number of moles reactive units in a compound. The formula we have used above is only used during acid-base titration. In some places in case of normality molarity might be given in that case same formula is used but normality will get replaced with molarity $ {M_1}{V_1}{\text{ }} = {\text{ }}{M_2}{V_2} $ .