For the equilibrium Al(OH)3( s )Al3++3OH( a q ) K sp=1.3×1033 Part A What is the
ID: 915325 • Letter: F
Question
For the equilibrium
Al(OH)3(s)Al3++3OH(aq)
Ksp=1.3×1033
Part A
What is the minimum pH at which Al(OH)3(s) will precipitate from a solution that is 6.5×102 M in Al3+?
Express your answer using two decimal places.
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Part B
A solution has [Al3+]= 6.5×102 M and [CH3COOH]= 1.00 M . What is the maximum quantity of NaCH3COO that can be added to 250.0 mL of this solution before precipitation of Al(OH)3(s)begins?
Express your answer using two significant figures.
For the equilibrium
Al(OH)3(s)Al3++3OH(aq)
Ksp=1.3×1033
Part A
What is the minimum pH at which Al(OH)3(s) will precipitate from a solution that is 6.5×102 M in Al3+?
Express your answer using two decimal places.
pH =SubmitMy AnswersGive Up
Part B
A solution has [Al3+]= 6.5×102 M and [CH3COOH]= 1.00 M . What is the maximum quantity of NaCH3COO that can be added to 250.0 mL of this solution before precipitation of Al(OH)3(s)begins?
Express your answer using two significant figures.
m = gExplanation / Answer
1)
Al(OH)3(s)Al3++3OH(aq)
Ksp=1.3×1033
Ksp = [Al+3][OH-]^3
if [Al+] = 6.5*10^-2
Then
1.3×1033 = ( 6.5*10^-2) (OH-)^3
[OH-] = 2.71*10^-11
pOH = -log(HO) = -log(2.71*10^-11 = 10.567
ph = 14-10.567
any pH > 3.433 will precipitate
Now that w ehave a pH
calculate
pH = pKa + log(A-/HA)
3.433 = 4.75 + log(A-/1)
Solve for [A-] =
[A-] )= 0.04819 M
V = 0.25 L so
mol A-. = M*V = 0.04819*0.25 = 0.0120475 mol of NaCH3COO
MW of Acetate = 82.0338
mass = mol*MW = 0.0120475*82.0338 = 0.988302 g of acetate