Question

14. (5 pts) Suppose the solubility in water of Al(OH)3 in g/L is 1.127 x 109. What is the K» for Al(OH)3? You must show the ionization reaction and your work!

Answer 1

ALCOH - Al3+ + 3040 Solubility = Convert the dividing the Mass of Molar Mass of 1.127 x 10-3912 (Given) Solubility in moles) 2 by Given Solubility with Molar ALCOH)₂ alcoH)2 = 26.98+ [16x3 + 1x3] 2 26.98 + 51 = 77.98 el mal in. = Now, Solubility molesti 1127x log xh mal 16 127 x 103 L 77.98 a 0.0144 x 10-3 molto = Now Ksp = [ Al 3+ ] [OH-] 3

Ksp = [ Al 3+ ] [OH-] Al3+ = s = 0.0144 x 10-3 moles ! [OH-] = 3s = 3 x 0.0144 x 10-3 L = 0.0432 X 10-3 moles L- Kspa Co.0144 X10-3) C0.0432 x 10-3) = 0.00 OOO U 60 x 10-12 Ksp = 116 X 10-18 Ksp, is the Mathematical product of its dissolved I ion concentration raised to the power of their stoi chometric Coefficients ..

General expression of Ksp My X₂ - M 27 + z XY* . Ksp= r M 2+] [ X Y -] 2. - of t Solubility Product o Molars O Molar Solubility the ions

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