Help with part c of this question please! (This is all the info provided to me)

Question

Help with part c of this question please! (This is all the info provided to me)

A solution contains 0.010 M Ba (aq) and 0.010 M Ag' (aq). (a) When titrated with chromate ion, which metal ion precipitates first? (b) What percentage of the initial concentration of this metal ion has been precipitated when the second metal ion begins to precipitate? Ksp BaCrO4-2.1 × 10 10 (e) If 0.010 M Ba2 (a) is mixed with 1.0 mM of another metal, M2* (aq), what is the maximum value of the Kp for MCrO4 that will allow the concentration of M2+ to be reduced by 99.9% before BaCrO4(s) begins to precipitate?

Explanation / Answer

(c) We need to determine the concentration of the metal ion M2+ in solution when BaCrO4 just begins to precipitate. Given the concentration of Ba2+ and the Ksp of BaCrO4, we determine the concentration of CrO42- when BaCrO4 just begins to precipitate.

We have

Ksp (BaCrO4) = [Ba2+][CrO42-]

====> 2.1*10-10 = (0.010)*[CrO42-]

====> [CrO42-] = (2.1*10-10)/(0.010) = 2.1*10-8

The concentration of CrO42- when BaCrO4 just begins to precipitate is 2.1*10-8 M. The said concentration of CrO42- reduces the concentration of M2+ by 99.9%; hence, the concentration of M2+ remaining in solution is just 0.1%, i.e., the concentration of M2+ in solution is (0.1%)*(1.0 mM) = (0.1/100)*(1.0 mM) = (0.1/100)*(1.0 mM)*(1 M/1000 mM) = 1.0*10-6 M (1 mM = 10-3 M).

We need to determine the Ksp of MCrO4 as

Ksp (MCrO4) = [M2+][CrO42-] = (1.0*10-6)*(2.1*10-8) = 2.1*10-14 (ans).

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