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# Answer to Question #126564 in Inorganic Chemistry for Ziyanda

Question #126564
1. Balance the following Redox reactions all occurring in basic medium
Fe(OH)3(s) + Cr(s) = Cr(OH)3(s) + Fe(OH)2(s)

2. Calculate the solubility of BaSO4 in pure water in g/L. Ksp BaSO4= 1.1×10^-10

3.The solubility product of Fe(OH)2 is 8×10^-16. Calculate the solubility of Fe(OH)2 in
(a) moles/l
(b)g/l
1
2020-07-17T05:40:57-0400

a) 3Fe(OH)3(s) + Cr(s) = Cr(OH)3(s) + 3Fe(OH)2(s)

b) let x be the mol/liter of solid  BaSO4.

Molar solubility = Ba2+ + SO4-2

In each case the charge is 2

therefore 1.1*10-10 = x2

thus x = square root in both cases that gives 1.05*10-5

therefore the solubilty of BaSO4 in pure water is 1.05*10-5

c) Molar solubility is given by

8×10^-16/1 *1/89.68 = 8.9 *10 -18

the molar solubility of  Fe(OH)2 i given by 4s3

thus the solubuty in moles = 4(8.9*1018)3

= 1.608 * 10-66 moles

in g =1.608*1066/89.86

1.78*10-68

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