Question

1.In gravimetric analysis, it is possible to use the common-ion effect to favor the production of solid precipitate. In this problem you will test this approach to see how effective it is for the gravimetric determination of calcium. Suppose the reaction flask in the gravimetric analysis experiment contains 175 mL of solution before filtering through the Gooch crucible. A precipitate of calcium oxalate monohydrate is collected in the crucible and dried, where:

CaC2O4 * H2O <--> Ca2^+ + C2O4^2- + H2O    Ksp=1.3x10^-8

The measured dry mass of CaC₂O₄·H₂O precipitate is 0.4324 g. In this problem you will determine the mass of Ca²⁺ that remains in the filtrate and is therefore unaccounted for in the precipitate. Specifically, please do the following:

(a) Determine the moles of oxalate ion in the 175-mL reaction flask before precipitation. Use the experimental information in the lab manual. For the purposes of this problem, assume the oxalate ion is totally deprotonated.

(b) Determine the equilibrium molarity of dissolved Ca²⁺ after the precipitation has occurred. [Hint: assume the precipitation first goes to completion; i.e., all the calcium ion from the original sample is in the precipitate. Then set up an ICE table to determine the amount (in moles/L) of the calcium ion that goes back into solution via solubility equilibrium subsequent to the precipitation. Note carefully that that the oxalate (“common ion”) is in excess before the precipitation.]

(c) Determine the mass (in g) of dissolved Ca²⁺ that remains in solution following filtration.

(d) Determine the mass (in g) of calcium ion in the solid precipitate.

(e) From your answers to (c) and (d), determine what percentage of the original calcium mass is lost to the filtrate.

(f) Repeat parts (b) through (e) under conditions in which the oxalate is not in excess; in other words, set the initial oxalate concentration in your ICE table to zero.

Answer 1

For the precipitation reaction,

(a) moles of C2O4^2- in solution before precipitation = 0.4324 g/128.097 g/mol = 0.0034 mol

(b) concentration of [Ca2+] If all has precipitated = 0.0034/0.175 = 0.019 M

ICE chart

                    CaC2O4 <==> Ca2+ + C2O4^2-

I                       0.019                -                -

C                        -x                   +x             +x

E                   (0.019 - x)          x                 x

Ksp = [Ca2+][C2O4^2-]

[Ca2+] = sq.rt.(1.3 x 10^-8) = 1.14 x 10^-4 M

(c) dissolved Ca2+ = 1.14 x 10^-4 x 0.175 x 40.08 = 0.0008 g

(d) mass of Ca in precipitate = 0.019 x 0.175 x 40.08 - 0.0008 = 0.0125 g

(e) Percentage of original calcium lost to the filtrate = 0.0008 x 100/0.133 = 0.60%