Ksp = 1.3 x 10^-33 for the precipitation of Al(OH)3 using 6.9lb of aluminum sulf

Question

Ksp = 1.3 x 10^-33 for the precipitation of Al(OH)3 using 6.9lb of aluminum sulfate, Al2(SO4)3

conversion of 6.9 lb into grams = 6.9* 453.6g. = 3129.84 g

conversion of gallons into liters = 4*1550*0.946 liters. = 5865.2 liters

solubility of Al2(SO4)3 will be 3129.84/(5865.2*342.15) = 0.00155921567 M

moles Al2(SO4)3 = 3129.84 g/ 342.13 g/mol=9.1475 moles  
moles Al3+ = 2 x 9.1475 = 18.295 moles  
[Al3+]= 18.295/5865.2 =0.00311926851 M

1.3 x 10^-33 = 0.00311926851[OH-]^3   

[OH-]= 7.5266205e-11 M

pOH = 10.1234011639
pH = 14 - pOH

pH = 3.87659

Explanation / Answer

The precipitation of aluminum hydroxide, Al(OH)3 (K,,-1.3 × 10 33), is sometimes used to purify water. Constants| Periodic Table Below are some potentially useful conversion factors. 1lb 1 gallon 1 quart 453.6 g = 4 quartts 0.946 L Part A At what pH will precipitation of Al(OH), begin f 6.90 lb of aluminum sulfate,AL( )3 is added to 1550 gallons of water (with a negligible change in volume)? Express your answer numerically to two decimal places. View Available Hint(s pH- Submit

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