My Math Forum Solubility equilibrium

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 February 28th, 2018, 09:58 AM #1 Newbie   Joined: Feb 2018 From: Iran Posts: 16 Thanks: 3 Solubility equilibrium When 3.06 g of SrCrO$_4$ is added to 500 mL of water, how many grams of SrCrO$_4$ precipitate? K$_{\text{sp}}$ = 2.2*10^-5 By the way, when I find molarity solubility by finding mole number of SrCrO$_4$ and divide it by 0.5, which is volume of solvent, I get 0.03, but when I find it using K$_{\text{sp}}$ S^2=2.2*10^-5 I get different answer. Why is this so? Last edited by skipjack; February 28th, 2018 at 10:50 AM.
 February 28th, 2018, 10:48 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,506 Thanks: 1741 What's the temperature of the water? Also, can you post your working in full, including any values you looked up?
February 28th, 2018, 08:19 PM   #3
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Quote:
 Originally Posted by skipjack What's the temperature of the water? Also, can you post your working in full, including any values you looked up?
I have tried to find molar solubility first, but I got stuck because in this case if I try to solve for molar solubility like this SrCrO$_4$â‡‹Sr+CrO$_4$ the molar mass is 203.61g/mol and mass is 3.06g, when we divide it by molar mass mole number is equal to 0.015mol and we find
molar solubility by dividing 0.015mol by 0.5L and it is equal to 0.03mol/L, but if I find the molar solubility in another way like SrCrO$_4$â‡‹Sr+CrO$_4$
s^2=2.2*10^-5 so s=4.69*10^-3mol/L the answers are different.

And temperature of water is not given.

Last edited by skipjack; February 28th, 2018 at 09:51 PM.

March 1st, 2018, 12:29 AM   #4
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In your first calculation, your arithmetic is correct, but the method relies on what assumption(s)?

Quote:
 Originally Posted by Elize . . . temperature of water is not given
The K$_\text{sp}$ is temperature dependent, so I assume its value was given in the question, as distinct from looked up by you.

Your calculation of the molar solubility as 4.69*10$^{-3}$mol/L is correct, so use it to calculate the mass of SrCrO$_4$ that is dissolved in 500mL of water, and hence the number of grams that "precipitate", given that 3.06g of SrCrO$_4$ was used.

March 1st, 2018, 02:57 AM   #5
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Quote:
 Originally Posted by Elize When 3.06 g of SrCrO$_4$ is added to 500 mL of water, how many grams of SrCrO$_4$ precipitate? K$_{\text{sp}}$ = 2.2*10^-5 By the way, when I find molarity solubility by finding mole number of SrCrO$_4$ and divide it by 0.5, which is volume of solvent, I get 0.03, but when I find it using K$_{\text{sp}}$ S^2=2.2*10^-5 I get different answer. Why is this so?
I make 3.06 g of strontium chromate to be 3.06/203.61 = 0.015 moles.

But strontium chromate is essentially insoluble so the calculation per litre is invalid.

That is .015 moles in 500 ml = .03 moles in 1L is invalid because it doesn't dissolve.

I said that it is 'essentially' insoluble.

However the solubility product indicates that a minute amount does dissolve.

Using this solubility product, you say you have calculated how much dissolves.

Did you get the fraction the right way up?

Hint write the equation down and correctly identify products/ reactants.

When you do this did you subtract this minute amount from the original 3.06g?

What do you think happens to the rest of the chromate?

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