Ncert Exercises & Solutions

Sulphuric acid reacts with sodium hydroxide as follows

$H_{2} {SO}_{4} + 2 NaOH \to {Na}_{2} {SO}_{4} + 2 H_{2} O$

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

1. 0.1 mol L^-1

2. 7.10 g

3. 0.025 mol L^-1

4. 3.55 g

(b, c)

For the reaction,

H_{2} {SO}_{4} + 2 NaOH \to {Na}_{2} {SO}_{4} + 2 H_{2} O

1L of 0.1 M H₂SO₄ contains=0.1 mole of H₂SO₄

1L of 0.1 M NaOH contains=0.1 mole of NaOH

According to the reaction, 1 mole of H₂SO₄ reacts with 2 moles of NaOH. Hence, 0.1 mole of NaOH will react with 0.05 mole of H₂SO₄ (and 0.05 mole of H₂SO₄ will be left unreacted), i.e., Hence, 0.1 mole of NaOH will produce 0.05 mole of Na₂SO₄.

Mass of {Na}_{2} {SO}_{4} = moles \times molar mass

= 0.5 \times (46 + 32 + 64) g

= 7.10 g

Volume of solution after mixing = 2 L

Since, only 0.05 mole of H₂SO₄ is left behind as NaOH completely used in the reaction. Therefore, molarity of the given solution is calculated from moles of H₂SO₄.

H_{2} {SO}_{4} left unreacted in the solution = 0.05 mole

∴ Molarity of the solution = \frac{0.05}{2} = 0.025 mol L^{- 1}

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