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When 1L of 0.1 M sulphuric acid solution...

When `1L` of `0.1 M` sulphuric acid solution is allowed to react with `1 L` of `0.1 M` sodium hydroxide then the molarity of sodium sulphate formed is
`(H_(2)SO_(4) + 2NaOH rarr Na_(2)SO_(4) + 2H_(2)O)`:

A

`0.1 M`

B

`0.05 M`

C

`0.025 M`

D

`0.2 M`

Text Solution

Verified by Experts

The correct Answer is:
C
`NaOH` is `LR`
`H_(2)SO_(4) + 2NaOH rarr Na_(2)SO_(4) + 2H_(2)O`
`[Na_(2)SO_(4)] = (0.05)/(2) = 0.025 M`
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