In a previous experiment, 0.50 g of Al was dissolved in 25.0 mL of 1.4 M KOH. Po

Question

In a previous experiment, 0.50 g of Al was dissolved in 25.0 mL of 1.4 M KOH. Potassium tetrahydroxoaluminate (KAl(OH)4) was neutralized by the addition of excess sulfuric acid. 5.0 mL of 9 M H2SO4 was added. Upon cooling the reaction solution, the target compoud, KAl(SO4)2 * 12H2O, crystallized. It is given that 14 g / 100 mL is the solubility reported for potassium aluminum sulfate dodecahydrate.

a) What is the molarity of the target compound?

b) Write the Ksp expression and calculate the Ksp constant value for this compound.

c) Under the reaction conditions given, about 0.30 M excess SO42- ion occurred in solution. Did this help or hinder precipitation? Reference the chemical principle.

Explanation / Answer

a)molar mass of target compound=KAl(SO4)2 * 12H2O=474.388g/mol

molarity of compound=14g/474.388 g/mol per 100ml=0.0295 mol/100ml=0.000295 mol/ml=0.000295*1000mol/L=0.295mol/L=0.295M

b)KAl(SO4)2 * 12H2O<--->K+   + Al3+ +2SO42-

ksp=[K+ ] [ Al3+ ] [SO42-]^2

let the solubility be S=0.295M

[K+ ] = [ Al3+ ]=S

[SO42-]=2S

ksp=S*S*S^2=S^4=(0.295M)^4=0.0257M

c)If [SO42-] =S+0.3

then the equilibrium shifts towards the product and the solubility becomes S+0.3

This hinders precipitation as more compound goes into the solution

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