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Offline bblue

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SOLUBILITY!!
« on: May 30, 2009, 03:00:53 PM »
Rank the following five salts in order of decreasing solubility, in terms of mass per unit volume.
(The most soluble gets rank 1, the least soluble gets rank 5.)

Ag3PO4     (Ksp = 1.80×10-18)
Ag2CrO4     (Ksp = 9.00×10-12)
Ca5(PO4)3F     (Ksp = 1.00×10-60)
Ca3(PO4)2     (Ksp = 1.30×10-32)
AgCl     (Ksp = 1.60×10-10)

So here is my work:

i first found the concentration of the ions:
Ag3PO4 : 1.60685E-5 M
Ag2CrO4 : 1.31037E-4 M
Ca5(PO4)3F : 6.109086E-8 M
Ca3(PO4)2 : 1.644765E-7 M
AgCl : 1.264911E-5 M

Then i used the concentrations and multiplied it by the molar mass and got the following solubilities [g / L]
Ag3PO4 : .0067259 g/L
Ag2CrO4 : .0434687 g/L
Ca5(PO4)3F : 3.0808E-5 g/L
Ca3(PO4)2 : 5.10167E-5 g/L
AgCl : .0081288 g/L

So the salt with the highest solubility would be the salt with the highest number:
3) Ag3PO4
1) Ag2CrO4
5) Ca5(PO4)3F
4) Ca3(PO4)2
2) AgCl

The rank that i have shown is incorrect  ???
and i dont know why
can someone PLEASE check my work
THANK YOU!!

Offline Borek

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Re: SOLUBILITY!!
« Reply #1 on: May 30, 2009, 04:35:27 PM »
i first found the concentration of the ions:
Ag3PO4 : 1.60685E-5 M
Ag2CrO4 : 1.31037E-4 M
Ca5(PO4)3F : 6.109086E-8 M
Ca3(PO4)2 : 1.644765E-7 M
AgCl : 1.264911E-5 M

Which ions?

How they are related to the concentration of the salt?
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Offline bblue

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Re: SOLUBILITY!!
« Reply #2 on: May 30, 2009, 09:24:33 PM »
well for the Ag3PO4 salt, this is how i found the concentration:

ksp = [3*x]^3 *

1.80×10-18 = 27(x^4)
x = 1.6E-5 M
x = [PO4]
and it would also equal to the concentration of the salt
wouldn't it?

Offline Borek

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Re: SOLUBILITY!!
« Reply #3 on: May 31, 2009, 04:25:54 AM »
So far so good. Show how you have dealt with Ca3(PO4)2.
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Offline bblue

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Re: SOLUBILITY!!
« Reply #4 on: May 31, 2009, 01:03:57 PM »
ok well here is my work for Ca3(PO4)2:

1.30E-32 = [3x]^3*[2x]^2
             = 108(x^5)
          x = 1.644765E-7 M
1.644765E-7 M * 310.17672 g = 5.10167E-5 g/L
is my work right?

Offline Borek

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Re: SOLUBILITY!!
« Reply #5 on: May 31, 2009, 02:38:47 PM »
No.

How many moles of Ca2+ per one mole of Ca3(PO4)2?
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Offline bblue

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Re: SOLUBILITY!!
« Reply #6 on: May 31, 2009, 03:20:04 PM »
wouldn't there be 3 moles for every 1 mole of Ca3(PO4)2??

Ca3(PO4)2  :rarrow: 3 Ca2+ + 2 PO42-

or is this eqn wrong, would Ca2+ actually exist as Ca22+ b/c the charges aren't really making any sense to me
I dont know how they got the 3 on the left side for the Ca

Offline Borek

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Re: SOLUBILITY!!
« Reply #7 on: May 31, 2009, 04:24:12 PM »
PO43-

Yes, 3 moles of Ca2+ per mole of Ca3(PO4)2. Now, if so - what is concentration of Ca3(PO4)2 for a given concentration of Ca2+?
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Offline bblue

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Re: SOLUBILITY!!
« Reply #8 on: May 31, 2009, 07:04:50 PM »
i THINK i understand what you are trying to say...

so the concentration i found is actually 3 times more than the concentration of my salt right?
but would i have to divide by 2 or would i have to divide by 3 ??
and what exactly did i find the concentration of ??

Offline Borek

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Re: SOLUBILITY!!
« Reply #9 on: May 31, 2009, 07:36:48 PM »
Argh, sorry - my mistake. You wrote

i first found the concentration of the ions:

and I was sure you did a classic mistake of not converting between ions/salt concentration, but your x was all the time salt concentration and your approach was correct.

Problem seems to be much simpler. Check your math for AgCl.
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Offline bblue

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Re: SOLUBILITY!!
« Reply #10 on: May 31, 2009, 08:10:27 PM »
i did the problem again and i got the same results...

AgCl  :rarrow: Cl- + Ag+

1.60E-10 =

1.60E-10 = x^2
x = 1.264911E-5 M

1.264911E-5 M * 143.321 g = .008128832 g/L

is there anything else that could be wrong with my calculations?

Offline Borek

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Re: SOLUBILITY!!
« Reply #11 on: June 01, 2009, 03:02:35 AM »
1.264911E-5 M * 143.321 g = .008128832 g/L

Check it again.
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Offline killer120

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Re: SOLUBILITY!!
« Reply #12 on: June 16, 2009, 11:36:22 PM »
Rank the following five salts in order of decreasing solubility, in terms of mass per unit volume.
(The most soluble gets rank 1, the least soluble gets rank 5.)

Ag3PO4     (Ksp = 1.80×10-18)
Ag2CrO4     (Ksp = 9.00×10-12)
Ca5(PO4)3F     (Ksp = 1.00×10-60)
Ca3(PO4)2     (Ksp = 1.30×10-32)
AgCl     (Ksp = 1.60×10-10)

So here is my work:

i first found the concentration of the ions:
Ag3PO4 : 1.60685E-5 M
Ag2CrO4 : 1.31037E-4 M
Ca5(PO4)3F : 6.109086E-8 M
Ca3(PO4)2 : 1.644765E-7 M
AgCl : 1.264911E-5 M

Then i used the concentrations and multiplied it by the molar mass and got the following solubilities [g / L]
Ag3PO4 : .0067259 g/L
Ag2CrO4 : .0434687 g/L
Ca5(PO4)3F : 3.0808E-5 g/L
Ca3(PO4)2 : 5.10167E-5 g/L
AgCl : .0081288 g/L

So the salt with the highest solubility would be the salt with the highest number:
3) Ag3PO4
1) Ag2CrO4
5) Ca5(PO4)3F
4) Ca3(PO4)2
2) AgCl

The rank that i have shown is incorrect  ???
and i dont know why
can someone PLEASE check my work
THANK YOU!!
hello....I'm a new member here!nice to meet you all!
i think the ranking is suppose like this :-
3) Ag3PO4
2) Ag2CrO4
5) Ca5(PO4)3F
4) Ca3(PO4)2
1) AgCl
is this the correct answer?
i think if we are trying to arrange the solubility of few salts,we should just compare the Ksp of the salts only,we are not suppose to calculate the concentration of each ion which exists in the salt.So,i think those equation is not necessary..
anyway,that's my opinion on the question.....i hope i can help someone there by reply this post! :)

Offline Borek

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Re: SOLUBILITY!!
« Reply #13 on: June 17, 2009, 03:01:55 AM »
i think if we are trying to arrange the solubility of few salts,we should just compare the Ksp of the salts only

Salt XY has solubility 0.01M - what is its Ksp?

Salt X2Y3 has 5 times higher solubility of 0.05M - what is its Ksp?

If you compare Ksp of both salts, will you get result telling you that second salt has greater solubility?
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Offline killer120

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Re: SOLUBILITY!!
« Reply #14 on: June 17, 2009, 04:21:33 AM »
i think if we are trying to arrange the solubility of few salts,we should just compare the Ksp of the salts only

Salt XY has solubility 0.01M - what is its Ksp?

Salt X2Y3 has 5 times higher solubility of 0.05M - what is its Ksp?

If you compare Ksp of both salts, will you get result telling you that second salt has greater solubility?
i already calculate the question you give me....OK below is my answer for your question :-
the Ksp for salt XY is [0.01][0.01]=0.0001M-2
the Ksp for salt X2Y3 is [2(0.05)]2[3(0.05)]3=0.00003375M-5

so,in the end the Ksp of salt XY is higher than Ksp for salt X2Y3 then i can know that salt XY is more soluble than salt X2Y3.
Am i right?
thanks for solving my misunderstanding during i learning this topic.....thank you again!now i know my mistake! ;D

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