A solution containing `10^(-3) M Sr(C1O_(4))_(2)` and `0.05 M KNO_(3)` was found to have only `75%` of its strontium in the uncomplexed `Sr^(2+)` form, the rest being `Sr(NO_(3))^(o+)`. Calcualate the `K_(1)` for complexation reaction:
`Sr^(2+) + NO_(3)^(Theta) rarr Sr (NO_(3))^(Theta)`