The specific gravity of `H_(2)SO_(4)` is `1.8 g// c c` and this solution is found to contain `98% H_(2)SO_(4)` by weight. `10c c` of this solution is mixed with `350 c c` of pure water. `25 mL `of this dil. `H_(2)SO_(4)` solution neutralises `500 mL` of `NaOH` solution. Then the `P^(H)` of `NaOH` solution is

Options

A. `12.398`
B. `1.602`
C. `12.699`
D. `12.301`

Answer

C

Explanation

Normaility of `H_(2)SO_(4) = ("Sp.gr" xx % xx 10)/("Ew.wt")`
`= (1.8 xx 98 xx 10)/(49) = 36`
on dilution with water, reultant
Normality of `H_(2)SO_(4) = (10 xx 36)/(360) = 1N`
`H_(2)SO_(4) + NaOH rarr Na_(2)SO_(4) + 2H_(2)O`
for neutralisation no. of mili equivalents of acid and base must be equal then
`N _(H_(2)SO_(4))V_(N_(2)SO_(4)) = N_(NaOH) V_(NaOH)`
`1 xx 25 = N_(NaOH) xx 500`
`N_(NaOH) = 1/20 = 0.5 = 5 xx 10^(-2)`
`P^(OH) = 2 - "log" 5 = 2 - 0.6990`
`p^(H)` of `NaOH = 12.6990`