If `2` litre of `9.8% w//w H_(2)SO_(4) (d = 1.5 g//mL)` solution is mixed with `3` litre of `1 M KOH` solution then the concentration ofv `H^(+)` if solution is acidic or concentration of `OH^(-)` if solution is basic in the final solution is:

Options

A. `0`
B. `(3)/(10) N`
C. `(3)/(5) N`
D. `(2)/(5) M`

Answer

C

Explanation

moles of `KOH` added are `= 3 xx 1`
`=3 mol`
`2 KOH + H_(2)SO_(4) rarr k_(2)SO_(4) + 2HOH`
`{:("initial moles",3,3,0,0),("moles after","reaction",3-3 3-1.5,1.5,3):}`
`= 0 = 1.5`
`:. [H^(+)] = (2 xx 1.5)/(2 + 3) = (3)/(5) M`