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

1 L of 0.1 M `H_(2)SO_(4)` contains =0.1 mol `H_(2)SO_(4)`
1 L of 0.1 M NaOH contains =0.1 mol NaOH According to the given equation, 1 mol of `H_(2)SO_(4)`
1 L of 0.1 M NaOH contains = 0.1 mol NaOH According to the given equation, 1 mol of `H_(2)SO_(4)` reacts with 2 mol of NaOH. Therefore, 0.1 mol of NaOH will react with 0.1/2 = 0.05 mol of `H_(2)SO_(4)` and 0.05 mol of `H_(2)SO_(4)` will remain unreacted. Therefore, NaOH is the limiting reagent.
`therefore 0.1` mol of NaOH will produce 0.05 mol `Na_(2)SO_(4)`
Molar mass of `Na_(2)SO_(4) = 2 xx 23 + 1 xx 32 + 4 xx 16 = 142`
Mass of `Na_(2)SO_(4)` formed `=0.05 xx 142 = 7.10`g
Volume of solution after mixing = 2L
`H_(2)SO_(4)` left unreacted = 0.05 mol
Molarity of solution `=0.05/2 = 0.025 mol L^(-1)`