Answer to Question #126872 in General Chemistry for Emma

Question

Does a precipitate form when 2.5 mLof 0.055M strontium nitrate isadded to 1.5 L of 0.011M zinc sulphate? Use calculations to justify your answer.

Answer 1

Zinc sulfate reacts with strontium nitrate, forming zinc nitrate and strontium sulfate.

Zinc Sulfate + Strontium Nitrate = Zinc Nitrate + strontium sulfate

During the reaction, zinc and strontium exchange the anions i.e. sulfate and nitrate ions.

ZnSO4 + Sr(NO3)2 = Zn(NO3)2 + Sr(SO4)

Whether a precipitate will be formed at the end of the reaction can be determined from a comparison of T.I.P (Trial Ion Product) and solubility product (K_SP)

since Zinc and sulfate ions exists in the ration of 1:1, molarity of sulfate ions = 0.011M

also, strontium and nitrate ions exists in the ration of 1:3, hence, molarity of strontium ions =0.0183M

molarity is given by = (moles*1000)/volume

Implying that the number of moles = (0.0183*2.5)/1000 = 4.575e^-5 moles of strontium ions present

the total resulting volume when the two solutions are added = 2.5+1500=1502.5mL

concentration of strontium ions now = 4.575e^-5/1.5025 = 3.045e^-5 mole/L

T.I.P of strontium and sulfate ions = (3.045e^-5)(0.011)

T.I.P = 3.3495 x 10−7

The K_sp for Sr(SO4) is 3.44 x 10⁻⁷ therefore T.I.P is less than K_sp so a ppt. will not be formed