You will need to read through the lab procedure and background information to so

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.

Explanation / Answer

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

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