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

Options

A. `"0.1 mol L"^(-1)`
B. 7.10 g
C. `"0.025 mol L"^(-1)`
D. 3.55 g

Answer

B, C

Explanation

Amount of `H_(2)SO_(4)` in 1L of 0.1 M solution = 9.8 g
Amount of NaOH in 1L of 0.1 M solution = 4.0 g
`underset(98g)(H_(2)SO_(4))+2underset(80g)(NaOH)rarrunderset(142g)(Na_(2)SO_(4))+2H_(2)O`
98 g of `H_(2)SO_(4)` need NaOH = 80 g
9.8 g of `H_(2)SO_(4)` need NaOH = 8.0 g
But NaOH actually available = 4.0 g
`:.` NaOH is the limiting reactant
Amount of `Na_(2)SO_(4)` formed
`=((142g))/((80g))xx(40g)=71g`
Molarity of `Na_(2)SO_(4)` solution (M)
`= (("7.1 g")//("142 g mol"^(-1)))/(2L)=0.025"mol L"^(-1)`.