Kaplan PS Section Test 2 #51

[steelersfan1243]

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1M NaOH is added to a solution containing 1m Ag+, 1M Al3+, 1M Mg2+, and 1M Mn2+. Given the solubility data shown below which of the following will precipitate first?
Ksp values:
AgOH 1.5x10-8
Al(OH)3 3.7x10-15
Mg(OH)2 1.2x10-11
Mn(OH)2 2.0x10-13

I calculated the molar solubility of each one and got Mg(OH)2 as the lowest, but the answer is A?

Any help would be appreciated, please and thank you.

EDIT: Changed in reference to azor ahai's comment

 

[azor ahai]

what is that data supposed to represent?

Ag is a transition metal, which aren't soluble with hydroxides.

 

[ctran019]

op, it is very unlikely the MgOH2 has the lowest molar solubility because you are dividing the KSP by 3 molecules so (x)(2x)2 vs the AgOH you are dividing by 2 molecules (x)(x) = you are comparing 10-3 vs 10-4.

 

[steelersfan1243]

Hmm, so let me explain my work then, I did this with a calculator just to save time, just to make sure the numbers were right:

AgOH-x2
Al(OH)3-27x4
MG(OH)2-4x3
Mn(OH)2-4x3

The molar solubilities I get are respectively 1,22E-4, 9.13E-6, 5.72E-5, and 1.46E-5. So this time I am getting B as the answer. I dont know if you're suppose to multiply by the given hydroxides in each compound, but still cant get A as an answer?

 

[Cawolf]

I don't think your math is correct.

I worked out the solubility of Al(OH)3 to be 1.08 e-4 M.

 

[Cawolf]

I worked out the first two using the Ksp values you provided. Maybe it will help you find an error.

http://i.imgur.com/QlHpczH.jpg

It is really large for some reason so I didn't embed.

I just noticed how crappy the chalkboard in this room is!

 

[steelersfan1243]

Yeah, I've been taking the cubed root an what not before dividing, oops....

Regardless, I am still not getting A as an answer? Thanks for all the help

 

[Cawolf]

(shorthand for the compounds)

Solubility of the hydroxides:

Ag = 1.2 e-4
Al = 1.08 e-4
Mg = 1.44 e-4
Mn = 3.68 e-5

I say the answer is Mn(OH)2.

I guess that doesn't fit your answer scheme but I am pretty sure I did the math right and Mn(II) hydroxide would be the least soluble and precipitate first.

Sorry.

 

[steelersfan1243]

Okay so I took another look at Kaplan's explanation and they are saying since the concentration of each metal M=1 molar the equilibrium equation reduces to Ksp=[OH]n, thus the saturation concentration of hydroxide ion in each case is [OH]=asqrtKsp

Why does this give a different answer than molar solubility?