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)`

Explanation

`Sr^(2+)` forms a very unstable complex with `NO_(3)^(Theta)`.
`{:(,Sr^(2+)+,NO_(3)^(Theta)rarr,Sr(NO_(3))^(Theta),),("Initial",rArr10^(-3)=0.001,0.05,0,),("used up",rArr0.00025,0.00025,,),("Ateq",rArr(0.001xx75)/(100),,,),(,=0.0075,0.05-0.00025,0.0025,),(,,~~0.05,,):}`
`K_(1) = ([Sr(NO_(3)^(o+))])/([Sr^(2+)][NO_(3)^(Theta)]) = (0.00025)/((0.00075)(0.05)) = 6.7`