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: 18,841 Thanks: 1564 
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  
Newbie Joined: Feb 2018 From: Iran Posts: 16 Thanks: 3  Quote:
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 
Global Moderator Joined: Dec 2006 Posts: 18,841 Thanks: 1564 
In your first calculation, your arithmetic is correct, but the method relies on what assumption(s)? 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  
Senior Member Joined: Jun 2015 From: England Posts: 797 Thanks: 234  Quote:
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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equilibrium, solubility 
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