Question

answer all parts (a and b) please!

Hydrogen sulfide (H2S) is a diprotic weak acid which can be used to selectively precipitate metal sulfide compounds from solution. The two deprotonation reactions of hydrogen sulfide are provided below along with their respective acid dissociation constants. H2S(ag) + H2O7) = HS (an) + H30+(aq) Kal = 1.0 * 10-7 HS (aq) + H2O) = 52 (ag) + H30 (29) K22=1 * 10-19 Imagine you have an aqueous solution buffered at pH 3 and containing the following species at the initial concentrations: [H2S] = 0.100 M and [In+1 = [Fe2+1 = 2 mM. (Ksp=4* 10-70 for In S and 8 * 10-19 for FeS.) (a) Which metal sulfide will precipitate out of solution, given the initial conditions? Provide calculations to support your answer quantitatively. (Not sure how to start? Here's a hint: what chemical species appears in both the deprotonation of hydrogen sulfide and the formation of both metal sulfides?) (b) You begin to add a strong base to your solution to increase its pH. At what pH will the second metal sulfide begin to precipitate?

Answer 1

Sol: IngS3 => a[In3+J+3[38] [Inst]2.[s?j! - 4X10-70 → Fes > [feat]+[53] [Beat]-[52] = $x1019 IS specified arbitarily that the separation Ps Quantitative When all the [In 3t except 1 part on 1000. That is when the concentration of cation it reduced to 1x104m By substituting this value in the expression of the product of solubility is obtained CS Scanned(1x10 4)* [-- 410-10 nner

[59] = 3.42x1621 m then it is calculated [52] necessary to start the prediction of the [Fetal in a solution 0.00am. precipitation begins just when the product of solubility is exceeded. The concentration of the {feta] is 0.0029 (0.002m) [5°-] = 8x10-19 [58] = 4 x 10t6m. These calculations show that the quantitative precipitation of In 3t ] occurs if [at] is greater than 3.42x152m. However, the Speta ] does not precipitate untill [52] is greater than 4x100m If we use the following equation based on the concentration of hydrogen sulfide in a saturated soln of gas is approximately oolm. . [Hot] a [s2] kika . And [Has] x 0.im and Ki=kal and kag = K2 equal to lxot and lxlota [H20t] [52]1xió?. 140079 chearing [san] 1X10 0. ta Camar

(30) - tematy na If we apply this equation to the above, we need the concentration range of Hidronio [t3 ot ] to conclude. Then to precipitate out the Inst] in a range is [H3 ot] = 1x1027 V [527 1x1027 = 0.0005M. V 3.42x1021m If [t3 ot] is maintained approximately blo 0.0005m and 0.002mm, in theory it can be quantitatively separate the [In 3t] from the rest] And pH = -log [t3o+] the concentration of Hydronium in the solution is . [+got] = 16PH = 103 Ezot] = 0.001M And [143043 cod 30.001 So, the first in precipate out is In3 +7 at concentration hydronium over 0.0005M. or Scanned with

o at pH in a concentration of hydronium equal to - 0.002mm then pH = -log[143 ot] = -log[0.002x103a] TpH = 5.70 cs Scared with CamScanner