Question

You will need to read through the lab procedure and background information to solve this question. Need help? See a TA or your instructor in office hours. CaCrO4 is a slightly soluble solid. In the lab, you find that addition of 357.73 µL of 1 M H2CrO4 to 7.924 mL of 0.3 M Ca(NO3)2 results in formation of a persistent precipitate. Addition of water (1.878 mL) just dissolves the precipitate. What is the experimental value of Ksp for CaCrO4 according to the information above? Provide your response to 4 digits after the decimal.

Answer 1

First write the balanced eq for formation of CaCrO4 :

Ca(NO3)2 (aq) + H2CrO4 (aq) ----> CaCrO4 (s) + 2HNO3 (aq)

Here volume of 1 M H2CrO4 = 357.73 µL

Or 357.73 3 x 10^-6 L

Number of moles of H2CrO4 which was added

= M x V

= 1 x 357.73 L = 0.000358

Moles of Ca(NO3)2 = 0.3 x 0.007924 = 0.00238

In this reaction H2CrO4 limiting reagent since both reactants react in 1:1 ratio and H2CrO4 moles is relatively less.

CaCrO4 moles formed = H2CrO4 moles reacted = 0.000358

total solution volume = 357.73 x 10^-6 L + 0.007924 L + 0.001878 L ( from water)

= 0.01016 L

Given precipitate dissolves at this volume

Solubility S = 0.000358 moles of CaCrO4 per 0.014 L

            = (0.000358 /0.01016) = 0.0352 moles /L or M

we have dissociation eq   CaCrO4 (s) <---> Ca2+ (aq) + CrO4^2- (aq)

Ksp = [Ca2+] [CrO4^2-]

Ksp = ( S) ( S) = S^2

= ( 0.0352)^2 = 1.2416*10^-3