Question

Sulphuric acid reacts with sodium hydroxide as follows:

$H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O$

When 1L of 0.1M sulphuric acid solution is allowed to react with 1L of 0.1M sodium hydroxide solution, the amount of sodium sulfate formed and its molarity in the solution obtained is

• A. 7.10 g
• B. 3.55 g
• C. $0.1mol{L}^{-1}$
• D. $0.025mol{L}^{-1}$

Answer 1

Answer: (A), (D)

$0.025mol{L}^{-1}$

Analysing the reaction.

$H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O$

No. of moles of solute = Molarity * Volume of solution

Molarity of $H_{2}S{O}_4$ = 0.1 M

0.1 moles of $H_{2}S{O}_4$ is present in 1L of solution

Molarity of NaOH = 0.1 M

Hence 0.1 moles of NaOH is present in 1L of solution

Solving for mass and concentration terms.

As 0.1 moles of NaOH will react with 0.05 moles of $H_{2}S{O}_4$. (As NaOH and $H_{2}S{O}_4$ react in ratio 2:1, because each molecule of $H_{2}S{O}_4$ gives 2 protons $\left(H^{+}\right)$ and each molecule of NaOH gives 1 hydroxide ion, in this acid-base reaction)

So NaOH will acts as a limiting reagent (because it will react completely)

From balanced equation,

$H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O$

As 2 moles of NaOH gives 1 mole of $N{a}_{2}S{O}_4$

$\therefore 10.1moleofNaOH=\frac{0.1}{2}=0.05mole$

So, mass of $N{a}_{2}S{O}_4=molesofN{a}_{2}S{O}_4\times \left(MolarMassofN{a}_{2}S{O}_4\right)$

$\to MassofN{a}_{2}S{O}_4=0.05\times (46+32+64)$

$\to MassofN{a}_{2}S{O}_4=7.10g$

As, volume of solution after mixing = 2L (i.e. 1L from each solution)

$MolarityofN{a}_{2}S{O}_4=\frac{molesofN{a}_{2}S{O}_4}{solutionvolume2}$

$=\frac{0.05}{2}$

$MolarityofN{a}_{2}S{O}_4=\frac{0.05}{2}=0.025mol{L}^{-1}$

Hence, from the above description it is clear that mass of $N{a}_{2}S{O}_4$ is 7.10 g.
and its molarity is $0.025mol{L}^{-1}$

So, option B and C are correct one.