25 mL of a solution of `Na_(2) CO_(3)` having a specific gravity of `1.25gmL^(-1)` required 32.9mL of a solution of HCI containing 109.5 g of the acid per litre for complete neutralization. Calculate the volume of `0.84NH_(2)SO_(4)` that will be completely neutralised by 125 g of `Na_(2) CO_(3)` solution.

Options

A. 470 mL
B. 360 ml
C. 580 mL
D. 250 mL

Answer

A

Explanation

`N _(HCl ) = ( 10 9. 5)/( 36. 5) = 3. `Since,` Na _(2) CO _(3)` is completely neutralized by HCl
`therefore Meq. of Na _(2) CO _(3) = Meq. of HCl , N xx 25 = 32. 9 xx 3 therefore N _( Na _(2) CO _(3)) = 3. 948`
Now, `Na _(2) CO _(3)` Fresh solution reacts with `H _(2) SO _(4),` Volume of `Na _(2) CO _(3)` solution `= (125)/(1.25) = 100 mL`
`therefore Meq. of H _(2) SO_(4) = Meq. of Na _(2) CO _(3), 0. 84 xx V = 100 xx 3. 948`
`therefore` Volume of `H _(2) SO _(4)` required `= 470 mL`