Question

Fill in the blank.
100 mL of 1M $\text{S}{{\text{O}}_{2}}\text{C}{{\text{l}}_{2}}$ aqueous solution is neutralised by _____mL of 4N $\text{NaOH}$.

Answer 1

Hint: The sulphuryl chloride is acidic whereas sodium hydroxide is basic in nature so when they react with each other they form salt and water and the reaction is known as neutralisation reaction. As they neutralise the acidic or basic effect of each other.

Complete step by step answer:
- In the given question we have to find the volume of sodium hydroxide which will be required to neutralise sulphuryl chloride.
- Firstly we will write the reaction between sulphuryl chloride and sodium hydroxide i.e.
\[\text{S}{{\text{O}}_{2}}\text{C}{{\text{l}}_{2}}\ \text{+ 4NaOH }\to \text{ N}{{\text{a}}_{2}}\text{S}{{\text{O}}_{4}}\text{ + 2NaCl + 2}{{\text{H}}_{2}}\text{O}\]

- Also, it is given that the molarity of sulphuryl chloride is 1 and the volume is 100 mL, so applying the formula of molarity we can calculate the moles of sulphuryl chloride i.e.
$\text{Molarity = }\dfrac{\text{No}\text{. of moles of solute }\times \text{ 1000}}{\text{Volume in mL}}$
$\text{1 = }\dfrac{\text{No}\text{. of moles of solute }\times \text{ 1000}}{100}$
$\text{No}\text{. of moles of solute = 0}\text{.01 moles}$.

- Now, the normality of sodium hydroxide is given. So, by applying the formula of normality we can calculate the volume of sodium hydroxide that is required for the neutralisation of sulphuryl chloride i.e.
$\text{Normality = }\dfrac{\text{No}\text{. of moles of solute }\times \text{ N-factor }\times \ 1000}{\text{Volume}}$

- Here, the normality is given 4N, N-factor for sodium hydroxide will be 1 because they give only 1 electron and also the charge on sodium and hydroxyl groups is +1 and -1 respectively.

- Whereas the no. of moles of solute is 0.4 moles because we know that for 1 molecule of sulphuryl chloride 4 molecules of sodium hydroxide are used so the no. of moles of sodium hydroxide will be:
$\text{0}\text{.1 }\times \text{ 4 = 0}\text{.4 moles}$.
$\text{4 = }\dfrac{\text{0}\text{.4 }\times \text{ 1 }\times \ 1000}{\text{Volume}}$
$\text{Volume = 100 mL}$.

Therefore, 100 mL of sodium hydroxide will be required to neutralise the sulphuryl chloride.

Note: In the given question one might get confused from the different concentration forms i.e. normality and molarity. Normality is the equivalent weight of the solution per litre whereas molarity is the number of moles of the solute per litre of solution.