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Sulphuric acid reacts with sodium hydrox...

Sulphuric acid reacts with sodium hydroxide as follows
`H_(2)SO_(4)+2NaOHrarrNa_(2)SO_(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 solphate formed and its molarity in the solution obtained is

A

`"0.1 mol L"^(-1)`

B

7.10 g

C

`"0.025 mol L"^(-1)`

D

3.55 g

Text Solution

Verified by Experts

The correct Answer is:
B, C
`"1 L of 0.1 M" H_(2)SO_(4)" contains = 0.1 mole of "H_(2)SO_(4)`
`"1 L of 0.1 M NaOH contains = 0.1 mole of NaOH"`
According to the given equation, 1 mole of `H_(2)SO_(4)` reacts with 2 moles of NaOH. Hence 0.1 mole of NaOH will react with 0.05 mole of `H_(2)SO_(4)` (and 0.05 mole of `H_(2)SO_(4)` will be left unreacted), i.e., NaOH is the limiting reactant.. 2 moles of NaOH produce 1 mole of `Na_(2)SO_(4)`. Hnece 0.1 mole of NaOH will produce 0.05 mole of `Na_(2)SO_(4)`
`=0.05xx(46+32+64)g=0.05xx142g=7.10g`
Volume of solution after mixing = 2 L
`H_(2)SO_(4)` left unreacted in the solution = 0.05 mole
`therefore" Molarity of the solution "=(0.05)/(2)=0.025" mol L"^(-1)`
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