Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Chemistry Olympiad and other competitions => Topic started by: Radu on April 04, 2014, 08:35:59 AM
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I have a problem: 0,01 moles of Ag3PO4 are placed in 1 L of pure water. Knowing ksAgOH=4*10-16, ksAg3PO4=2.7*10-18, Ka1=10-2, ka2=10-7, ka3=10-2 (for H3PO4), calculate:
a) the constant of the reaction: Ag3PO4+ 3H2O ::equil:: 3AgOH + H3PO4.
b) find pH, [PO43-], and [Ag+] at equilibrium in the resulting solution
c) ignoring the hydrolisis of Ag+, calculate the pH of the solution.
Here's my approach:
a) I obtain easily k=4.2*107.
b) Having this big constant for the hydrolisis of the salt, I can assume that all silver phosphate has dissolved, and, at equilibrium, that : [Ag+]=ksAgOH/[OH-]., and CH3PO4≈0.01M. Given we expect to have a basic solution, I do the charge balance: [PO43-]*3+ [HPO42-]*2+[HO-]=[H3O+] + ksAgOH/[HO-]. Multiplying the relation with [HO-], I obtain: some terms + [HO-]2= Kw + ksAgOH≈kw. some terms=kw-[HO-]2, which means that [OH-]<10-7 ( acidic medium) Is this actually correct? Does silver phosphate hydrolisis create acidity? The correct answers are [Ag+]=6.5*10-14M, [PO43-]=3.8*10-3, pH=11.792. If you look at these, you'll notice ksAgOH has been reached, and that all Ag3PO4 has been dissolved
c) it is very similar to the phosphoric acid problem(IChO-see "Back with the phosphoric acid" topic), but applying the same method I obtain pH= 9.8. They obtain pH=10.62. One thing that strikes is that this pH, which is achieved theoretically, by neglecting Ag+ hydrolisis, is smaller than the one where Ag+ lowers the pH by forming the insoluble hydroxide). What might be wrong, and how can this problem be treated?
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I have a problem: 0,01 moles of Ag3PO4 are placed in 1 L of pure water. Knowing ksAgOH=4*10-16, ksAg3PO4=2.7*10-18, Ka1=10-2, ka2=10-7, ka3=10-2 (for H3PO4), calculate:
a) the constant of the reaction: Ag3PO4+ 3H2O ::equil:: 3AgOH + H3PO4.
b) find pH, [PO43-], and [Ag+] at equilibrium in the resulting solution
c) ignoring the hydrolisis of Ag+, calculate the pH of the solution.
Here's my approach:
a) I obtain easily k=4.2*107.
b) Having this big constant for the hydrolisis of the salt, I can assume that all silver phosphate has dissolved, and, at equilibrium, that : [Ag+]=ksAgOH/[OH-]., and CH3PO4≈0.01M. Given we expect to have a basic solution, I do the charge balance: [PO43-]*3+ [HPO42-]*2+[HO-]=[H3O+] + ksAgOH/[HO-]. Multiplying the relation with [HO-], I obtain: some terms + [HO-]2= Kw + ksAgOH≈kw. some terms=kw-[HO-]2, which means that [OH-]<10-7 ( acidic medium) Is this actually correct? Does silver phosphate hydrolisis create acidity? The correct answers are [Ag+]=6.5*10-14M, [PO43-]=3.8*10-3, pH=11.792. If you look at these, you'll notice ksAgOH has been reached, and that all Ag3PO4 has been dissolved
c) it is very similar to the phosphoric acid problem(IChO-see "Back with the phosphoric acid" topic), but applying the same method I obtain pH= 9.8. They obtain pH=10.62. One thing that strikes is that this pH, which is achieved theoretically, by neglecting Ag+ hydrolisis, is smaller than the one where Ag+ lowers the pH by forming the insoluble hydroxide). What might be wrong, and how can this problem be treated?
You mean Ag3PO4 (s) + 3H2O (l) ::equil:: 3AgOH (s) + H3PO4 (aq)? I want to be careful to avoid ambiguity.
If the constant is that high for this equilibrium, it is implied that most PO43- ends up as H3PO4 and most Ag+ as AgOH. So how does the answer give a decently high PO43-? Of course it is implied that Ksp(AgOH) is reached anyway, otherwise your equilibrium constant does not hold since this equilibrium: Ag3PO4 (s) + 3H2O (l) ::equil:: 3AgOH (s) + H3PO4 (aq) does not exist unless AgOH saturation has been reached.
Where did you get the problem?
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Is it from icho?
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Look up Woodward's Nobel lecture from 1966 I think it was published in Science.