1.In gravimetric analysis, it is possible to use the common-ion effect to favor the production of solid...

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Chemistry

1.In gravimetric analysis, it ispossible to use the common-ion effect to favor the production ofsolid precipitate. In this problem you will test this approach tosee how effective it is for the gravimetric determination ofcalcium. Suppose the reaction flask in the gravimetric analysisexperiment contains 175 mL of solution before filtering through theGooch crucible. A precipitate of calcium oxalate monohydrate iscollected in the crucible and dried, where:

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

The measured dry mass ofCaC2O4·H2O precipitate is 0.4324g. In this problem you will determine the mass ofCa2+ that remains in the filtrate and is thereforeunaccounted for in the precipitate. Specifically, pleasedo the following:

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

(b) Determine the equilibrium molarity of dissolvedCa2+ after the precipitation has occurred. [Hint: assumethe precipitation first goes to completion; i.e., all the calciumion from the original sample is in the precipitate. Then set up anICE table to determine the amount (in moles/L) of the calcium ionthat goes back into solution via solubility equilibriumsubsequent to the precipitation. Notecarefully that that the oxalate (“common ion”) is inexcess before the precipitation.]

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

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

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

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

Answer & Explanation Solved by verified expert
3.7 Ratings (380 Votes)

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%


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