Question

A 50.0 mL sample of 0.0645 M AgNO3(aq) is added to 50.0 mL of 0.100 M NaIO3(aq). Calculate the [Ag+] at equilibrium in the resulting solution. The Ksp value for AgIO3(s) is 3.17 × 10-8.

[Ag⁺] =____ mol/L

Answer 1

If we add 50 ml of 0.0645 M AgNO3 in 50 ml of 0.1 M NaIO3 the total volume will be 50ml+50ml = 100ml

millimoles of Ag+ in AgNO3 = millmoles of AgNO3 = molarity*volume in ml

= 0.0645M*50ml = 3.225 millimoles

So, the new concentration of Ag+ just after adding the solutions = millimoles/volume in ml

= 3.225 millimoles/100 ml = 0.03225 M

Also,

millimoles of IO3- in NaIO3 = millmoles of NaIO3 = molarity*volume in ml

= 0.1M*50ml = 5 millimoles

So, the new concentration of IO3- just after adding the solutions = millimoles/volume in ml

= 5 millimoles/100 ml = 0.05 M

now we check the ionic product of Ag+ and IO3-

[Ag+]*[IO3-] = 0.03225 * 0.05 = 0.0016125 which is greater than the Ksp value of AgIO3(s)

which is 3.17 × 10^-8

So, Ag+ and IO3- will get consumed to form precipitate AgIO3(s)

Ag+ + IO3- -> AgIO3

I 0.03225 0.05 0  

C -x -x +x

E 0.03225-x 0.05-x x

K = 1/Ksp = 1/3.17 × 10^-8 = 1/[Ag+][IO3-]

So, [Ag+][IO3-] = 3.17 × 10^-8

[0.003225-x]*[0.05-x] = 3.17*10^-8

0.0016125 - 0.03225x - 0.05x + x² = 3.17*10^-8

x²- 0.08225x + 0.0016125 - 3.17*10-8 = 0

so, x = (0.08225 - sqrt(0.08225² - 4*1*(0.0016125 - 3.17*10-8 ) ))/2

so, x = (0.08225 - 0.01775357147)/2 = 0.03224821426 M

so, molarity of Ag+ at equilibrium = 0.03225 M- 0.03224821426 M

= 0.00000178573 M

= 1.786 * 10^-6 M