Question

What is the strength in gram per litre of a solution of$\text{ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }$, 12 mL of which is neutralised by 15 mL of $\dfrac{\text{N}}{\text{10}}\text{ }$$\text{ NaOH }$solution?
A) $\text{ 6}\text{.125 }$
B) $\text{ 8}\text{.215 }$
C) $\text{ 6}\text{.625 }$
D) $\text{ 8}\text{.125 }$

Answer 1

Hint: At a neutralisation point, the milliequivalent of acids are always equal to the milliequivalent of the base. That is if $\text{ }{{\text{V}}_{\text{1}}}\text{ }$ the volume of $\text{ }{{\text{N}}_{\text{1}}}\text{ }$ the strength of acid reacts with the $\text{ }{{\text{V}}_{2}}\text{ }$ volume of the base having $\text{ }{{\text{N}}_{2}}\text{ }$ strength. Then,
${{\text{N}}_{\text{1}}}{{\text{V}}_{\text{1}}}\text{ = }{{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}}\text{ }$
Where N is the normality of acid or base and V is the volume of acid or base.

Complete step by step solution:
The reaction at which the acid reacts with the base is called the acid-base reaction. The point at which acid completely neutralizes the base to form salt and water is known as the neutralisation reaction. Let's consider neutralisation reaction between strong acid sulphuric acid $\text{ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }$ and strong base sodium hydroxide $\text{ NaOH }$ . The reaction between the sulphuric acid and sodium hydroxide is as shown below,
$\text{ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ + 2NaOH }\to \text{ N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ + 2}{{\text{H}}_{\text{2}}}\text{O }$
At the point of neutralisation, the miliequivalent of sulphuric acid $\text{ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }$is equal to the miliequivalent of sodium hydroxide $\text{ NaOH }$. That is,
$\text{ milliequivalent of }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4 }}}\text{= milliequivalent of NaOH }$
Miliequivalent is expressed as the product of normality of solution and volume of solution. Consider that$\text{ }{{\text{V}}_{\text{1}}}\text{ }$ volume of $\text{ }{{\text{N}}_{\text{1}}}\text{ }$ the strength of sulphuric acid reacts with the $\text{ }{{\text{V}}_{2}}\text{ }$ volume of $\text{ }{{\text{N}}_{2}}\text{ }$ the strength of sodium hydroxide. Then at point of neutralisation,
${{\text{N}}_{\text{1}}}{{\text{V}}_{\text{1}}}\text{ = }{{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}}\text{ }$ (1)
We have given that,
The volume of sulphuric acid is$\text{ }{{\text{V}}_{\text{1}}}\text{ = 12 mL }$
The volume of sodium hydroxide utilized to neutralized sulphuric acid is $\text{ }{{\text{V}}_{2}}\text{ = 15 mL }$
Strength of sodium hydroxide is $\text{ }{{\text{N}}_{\text{2}}}\text{= }\dfrac{1}{\text{10}}\text{ N }$
Let's substitute these values in equation (1). we have,
$\text{ }{{\text{N}}_{\text{1}}}\left( \text{12 mL } \right)\text{ = }\left( \dfrac{1}{\text{10}} \right)\left( \text{15 mL } \right)\text{ }$
On further solving we get,
$\begin{align}
& \text{ }{{\text{N}}_{\text{1}}}\text{= }\left( \dfrac{1}{\text{10}} \right)\dfrac{\left( \text{15 mL } \right)}{\left( \text{12 mL } \right)}\text{ } \\
& \Rightarrow {{\text{N}}_{\text{1}}}\text{ = 0}\text{.125 N } \\
\end{align}$
We are interested to determine the strength of an acid in terms of a gram per equivalent. The normality of the solution is written as follows,
$\text{ Normality = }\dfrac{\text{Equivalent weight}}{\text{Volume}}\text{ }$
Here, the equivalent weight of sulphuric acid is calculated by taking the ratio of molecular weight to the valence factor. Sulphuric acid can donate two protons .Thus it has a valence factor equal to 2.Thus equivalent weight of sulphuric acid is,
$\text{ Equivalent weight of }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ = }\dfrac{\text{MW}}{\text{2}}\text{ = }\dfrac{98g}{2}\text{ = 49 g }$
Since the molecular weight of sulphuric acid is 98 grams.
We have to find the gram per litre of the solution .Thus strength of an acid is,
$\text{ Strength (in gram }{{\text{L}}^{\text{-1}}}\text{) = N }\times \text{ Equivalent weight = 0}\text{.125 }\times \text{ 49 = 6}\text{.125 g}{{\text{L}}^{-1}}\text{ }$
Thus, the strength of sulphuric acid in gram per litre of the solution to neutralize the sodium hydroxide is equal $\text{ 6}\text{.125 g}{{\text{L}}^{-1}}\text{ }$ .

Hence, (A) is the correct option.

Note: Note that, the no of equivalent is dependent on the characteristic properties of substance. This is also called a valence factor. For acids, valence factor for acid=basicity=removal no.of proton (${{\text{H}}^{\text{+}}}$) For bases, valence factor for base=acidity=removal no.of hydroxide (\[O{{H}^{-}}\]).