1. (60%) Gibbsite (Al(OH) 3 ) is one of the commonest forms of aluminum found in

Question

1. (60%) Gibbsite (Al(OH)3) is one of the commonest forms of aluminum found in the natural environment and is a product of weathering alumino-silicates. Consider the dissociation of the solid gibbsite (log K = 7.74 at 25oC):

Al(OH)3(s) + 3H+ Al3+ + 3H2O

and the formation of the complex Al(SO4)+ which has a stability constant (Kstab) also known as a formation constant Kf = 103.01 (at 25oC):

Al3+ + SO42- AlSO4+

(a) In an aqueous environment at pH 4 and [SO42-] = 10-3 M, calculate the concentration (in moles) of total Al in equilibrium with Al(OH)3(s) (gibbsite).

Explanation / Answer

given in second reaction

Al3+ + SO42- = AlSO4+

kf= concentration of products/ concentration of reactants

kf= [AlSO4+]/[SO42-][Al3+]

given [SO42-]= 10^-3 and [AlSO4-]=1 since solid phase

103.01= [1]/10^-3 *[Al3+]

[Al3+]= 10^3/103.01....................(1)

now consider first reaction

k=([Al3+][H2O]^2)/([Al(OH)3]*[H+]^3)

since Al(OH) 3 is solid its concentration is =1 and similarly since H2O is liquid its concentrationn is also =1

take log both side

logk=3log[Al3+](-log[H+])

given logk- 7.74 and we know -log[H]= ph= 4

7.74/3*4=log[Al3+]

0.645= log[Al3+]

[Al3+]= e^0.645=1.90 moles

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