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Question

Sulphuric acid reacts with sodium hydroxide as follows

H2SO4+2NaOHNa2SO4+2H2O

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

(a) 0.1 mol L1 (b) 7.10 g

(c) 0.025 mol L1 (d) 3.55 g

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Solution

(b)

For the reaction, H2SO4+2NaOHNa2SO4+2H2O

1L of 0.1 M MH2SO4 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., NaOH is the limiting reactant. Since, 2 moles of NaOH produce 1 mole of Na2SO4.

Hence, 0.1 mole of NaOH will produce 0.05 mole of Na2SO4.

Mass of Na2SO4=moles×molar mass

=0.5×(46+32+64)g

= 7.10 g

Volume of solution after mixing = 2L

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.

H2SO4 left unreacted in the solution = 0.05 mole

Molarity of the solution =0.052=0.025mol L1


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