If a solution of 20mL of 0.050M K + is added to 80mL of 0.50M ClO4 - will a precipitate form and what is the value of Qsp? For KClO4, Ksp = 1.07 x 10-2

Q< K hence a precipitate will not form.

Explanation:

First convert the concentration to molL-1

For number of moles of K^ + = 0.05*20/1000 = 1*10^-3 moles

If we have 1*10^-3 moles in 20cm3

Then in 1000cm^3 we have 1*10^-3 moles*1000/20 = 0.05 M

For ClO4^- = 0.50*80/1000 = 0.04 moles

If we have 0.04 moles in 80cm3

Then in 1000cm^3 we have 0.04 moles*1000/80 = 0.5 M

Q = [K^+] [ClO4^-]

Q = [0.05] [0.5]

Q = 0.025 = 2.5*10-2

Q< K hence a precipitate will not form.