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

Question

1·(60%) Gibbsite(Al(OH)3) 1s 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 25°C) and the formation of the complex Al(SO4) which has a stability constant (Kstab) also known as a formation constant K,-10 (at 25°C) 3.01 (a) In an aqueous environment at pH 4 and [SO4_-10% M, calculate the concentration (in moles) of total Al in equilibrium with Al(OH)3s) (gibbsite) (b) Calculate the ratio [AISO4+] [AP]. Assume no other complexes are being formed (c) The formation of the complex Al(SO4)2 has a stability constant, Kstab, of 10490 (at 25'C) Using the same conditions as in (a), calculate the ratio [Al(SO4)2]/ [Al" (d) Assume the same conditions as above, except that [SO42-] = 0.1 M, calculate the ratios [AISO.T [Al3+1 and [Al(SO4)2-]/ [A (e) Calculate the concentration (in M) of total Al in equilibrium with Al(OH)3(s) (gibbsite) when [SO4 1-0.1 M 2- Briefly discuss the effect that increasing the ligand(SO42-) concentration has on the formation of Al complexes and on the total [Al] dissolved in the water.

Explanation / Answer

answer 1: 1 mole of gibbsite give 1 mole of Al3+ so 0.6 moles give 0.6 moles

by simple formula

logK=1.8(log [Al3+]^0.6)*(ph)

logk/(ph*1.8)= log(Al3+)

7.74/4*1.8=1.051= log(Al3+)

e^1.051= moles of Al3+

2.74 moles of Al+

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