Question

What is the strength in g per litre of a solution of sulphuric acid, $\text{12 mL}$ of which neutralise $\text{15 mL}$ of $\dfrac{N}{10}$ sodium hydroxide solution?

Answer 1

Hint: In the given question, the given data is the volume of sulphuric acid which neutralises a certain volume of sodium hydroxide which has a tenth of its normality. The strength of sulphuric acid is to be found. We must know the fundamental definition of strength of an acid. Acid strength signifies the tendency of any acid to dissociate into a proton and an anion. Keeping this in mind, we can easily solve the given question.

Complete step by step answer:
First we have to write the balanced equation. So, here is the equation
$2NaOH+{{H}_{2}}S{{O}_{4}}\to N{{a}_{2}}S{{O}_{4}}+2{{H}_{2}}O$

${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$
Then we have to calculate the equivalents of NaOH.
Now the equivalents are as follows
0.015 L NaOH $\times $ $\dfrac{\text{0}\text{.1 equivalent NaOH}}{\text{1 L NaOH}}$ = 0.0015 equivalent NaOH.
Now we have to calculate the equivalents of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$. We should know that 1 equivalent of anything is 1 equivalent of anything else in case of a reaction. Hence, there are 0.0015 equivalent of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$.
Now,
The mass of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ is = 0.0015 equivalent of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$$\times $ $\dfrac{\text{1mol }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}}{\text{equivalent }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}}$= 0.00075 mol ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$.
Now we have to calculate the mass of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ in the sample.
So mass of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ = 0.0007 mol ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$$\times $ $\dfrac{\text{98}\text{.08g }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}}{\text{12mL solution}}$= 0.074 g of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$.
Now the mass of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ in 1 litre of solution.
Mass of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$= 1000 mL solution $\times $ $\dfrac{\text{0}\text{.074g }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}}{\text{12mL solution}}$= 6.1 g of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$.

Note: Sulphuric acid is an example of a strong acid. A weak acid is only partially dissociated while a strong acid is fully dissociated. The strength of a weak acid is determined by its acid dissociation constant.