Will a precipitate form?

Question

The solubility of a hydroxide M(OH)_2 is 0.0090 g per liter. The molecular weight of M(OH)_2 is 150. If 1.0 gram of MSO_4 is dissolved in 500 mL of water buffered at pH 8.50, will a precipitate form? (H= 1.009 g/mol, O= 16.00 g/mol, S=32.06 g/mol)

Answer 1

now let's find the Ksp of M(OH)2:

first comes molarity:

0.0090 grams / litre @ 150 grams /mole = 6.0e-5 Molar M(OH)2

which has released --> 1 M+2  & 2 OH-

at molarities of  6.0e-5 Molar & 12.0e-5 Molar

Ksp = [M+2] [OH]^2

Ksp =  [ 6.0e-5] [ 12 e-5]^2

Ksp = 8.6e-13

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now we will try to find what is in this solution:

if the molecular weight of M(OH)_2 is 150 , then the atomic weight of M is 116 g/mole ( because two OH's weigh 34)

If 1.0 gram of MSO_4 is dissolved in 500 mL ,

and if the molar mass of MSO4 is 116 + 32 + 64 = 212 g/mole , then

let's find the moles:

1 gram @ 212 g/mole = 0.004717 moles of MnSO4

now let's find the molarity of the MSO4:

0.004717 moles / 0.500 litres = 0.009434 Molar MnSO4

however it is a 1:1 ratio between MSO4 & Mn , so the molarity of the Mn is also 0.009434 Molar

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now let's find the OH- molarity:

@ a pH = 8.50,  there is a pOH of 5.50 ,...

a pOH of 5.50 .... equals a [OH-] of 3.16e-6 Molar

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so let's find the Q

Q = [M+2] [OH-]2

Q = [0.009434 M] [3.16e-6 M]2

Q = 9.42e-14

remembering that Ksp = 8.6e-13 for a saturated solution, Q is lee than K, the solutionis unsaturated

your answer is : no precipitate