Sulphuric acid reacts with sodium hydroxide as follows
$H_{2} {SO}_{4} + 2 NaOH ⟶ {Na}_{2} {SO}_{4} + 2 H_{2} O$
When 1 L 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 are

Question

Sulphuric acid reacts with sodium hydroxide as follows

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

When 1 L 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 are

Infinity Learn · Accepted Answer

Moles of $$H2SO4=1L$$×$$0.1M=0.1molH2SO41$$ $$mol+2$$ $$NaOH2$$ mol→$$Na2SO41$$ $$mol+2H2O$$  $$0.1$$ mol   $$0.1$$ mol       $$0.05mol$$∴ Molarity mass of $$Na2SO4=0.05$$×$$142=7.1g$$∴ Molar mass of $$Na2SO4=142$$ Molarity $$=$$ moles of $$Na2SO4$$ volume of solution $$=0.051+1=0.025$$ mol L−$$1$$

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Sulphuric acid reacts with sodium hydroxide as follows H2SO4+2NaOH⟶Na2SO4+2H2OWhen 1 L 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 are

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