`17.4%` (wt.`//`vol.) `K_(2)(SO_(4))` solution at `27^(@)C` isotonic to `5.85%` (wt.`//`vol.) `NaCl` solution at `27^(@)C`. IF `NaCl` is `100%` ionised, what is `%` ionisation of `K_(2)SO_(4)` in aq. solution?

Answer

`50%K_(2)SO_(4)`

Explanation

For isotonic solutions, `i_(1)C_(1)=i_(2)C_(2)`
for `K_(2)SO_(4)` and `NaCl, i_(1)xx(17.4xx1000)/(174xx100)=(2xx5.85xx1000)(58.5xx100)`
so `i_(1)=2`
Now `K_(2)SO_(4) rarr2K^(+)+SO_(4^(2-))`
At equilibrium `1-x 2x x`
or `1+2x=2`
or `x=1//2=0.5`
`therefore 50%` ionisation.