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The specific gravity of H(2)SO(4) is 1.8...

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

A

`12.398`

B

`1.602`

C

`12.699`

D

`12.301`

Text Solution

Verified by Experts

The correct Answer is:
C
Noramility of `H_(2)SO_(4)=(Sp. Gr xx%10)/(Eq.wt)`
`=(1.8xx98xx10)/(49)=36`
Normality of `H_(2)SO_(4)=(10x36)/(360)=1N`
`H_(2)SO_(4) to NaOH to 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)`
`xx2=N_(NaOH)xx500`
`N_(NaOH)=(1)/(20)=0.5=5xx10^(-2)`
`P^(OH)=2-log5=2-0.6990`
`P^(H)` of `NaOH=12.6990`
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