Answer to Question #126564 in Inorganic Chemistry for Ziyanda

Question

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

Answer 1

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⁺ + SO_4-2

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