Question

A solution containing sr(no3)2 is mixed with a solution of Na2CrO4 to form a solution that is 0.0105 m in Sr(no3)2 And 0.0055 m in Na2Cro4

Ksp(SrCrO4) =7.6*10^-5

what will happen once these solutions are mixed?

a) my Q reaction non equilibrium quotient us=

B) circle 2 possible choices for products based in W in a)

Yes precipitate

No precipitate

Answer 1

0.0105M solution of strontium nitrate contains 1eq of Sr²⁺ and 2eq of NO₃^- while 0.0055M sodium chromate will contain 2eq sodium ion and 1eq chromate ion. Therefore, when the two solutions are mixed, the quantity of strontium chromate formed will be dictated by the lowest quantity of ions available which is chromate. Thus, we can say that with sodium chromate as the limiting reagent, the solution will have 0.0055M strontium chromate.

The solubility product serves as a measure of solubility and insolubility of a substance in a given solvent. From this parameter, the maximum concentration of the substance that can be in solution phase, without precipitation, known as "molar solubility" can ber determined in terms of molarity.

The Ksp of strontium chromate is expressed as K_sp = [Sr²⁺][CrO₄^2-]/[SrCrO₄]. For molar solubility, considering a total concentration of 1M of strontium chromate, the equation becomes K_sp = [Sr²⁺][CrO₄^2-]. Since 1M solution of the salt dissociates into 1M Sr²⁺ and CrO₄^2- respectively, both parameters can be taken as a single constant "x". This modifies the equation as K_sp = x² => sq.rt of 7.6x10^-5 = x. This gives the maximum concentration of strontium and chromate ions possible in water in molarity, from strontrium chromate as 8.7178x10^-3M. So, it is now known that 8.7178x10^-3M of the salt can be present in water. From the concentration of the salt given, we find that there is 5.5x10^-3M in the resultant solution. As this quantity is considerably less than the maximum quantity of the salt soluble, no precipitation will be observed.