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Topic: Practice problem 3  (Read 5159 times)

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Offline Rutherford

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Practice problem 3
« on: April 24, 2014, 07:30:37 AM »
10.0 g of white phosphorus were chlorinated with elementary chlorine. At the end of the reaction no unreacted phosphorus left. The products mixture was hydrolyzed. 10 ml of this solution was titrated with 0.25 M potassium hydroxide (phenolphthalein was used). The consumption was 8.7 ml. Another 10 ml sample was titrated with potassium hydroxide of same molarity, but using methyl orange as indicator. Consumption of potassium hydroxide in this case was 7.1 ml. Calculate the mass of chlorine used in the experiment.
Hint: pKa(phenolphthalein) = 9.5, pKa(methyl orange) = 3.5

Offline SinkingTako

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Re: Practice problem 3
« Reply #1 on: April 24, 2014, 08:24:29 AM »
Okay, I made quite a few assumptions here:
Assume that all the chlorine reacted with the phosphorus.
And for the PCl3, PCl5 formed, It fully hydrolyse to H3PO3 and H3PO4.

So
PCl3 + H2O :rarrow: H3PO3 +3HCl (pKa = 1.3, 6.7)
PCl5 + H2O :rarrow: H3PO4 +5HCl (pKa = 2.15, 7.20, 12.3)

Looking at the pKa values, in methyl orange, the phosphurous species present will be H2PO4- and H2PO3- predominantly.

In pheno, the phosphurous species present will be HPO42- and HPO32- predominantly.

So in the 10ml reacted, let there be x mol H3PO3 and y mol H3PO4

4x + 6y = 0.0071 x 0.25
5x + 7y = 0.0087 x 0.25

solving, x= 3.125 x 10-4, y= 8.75 x 10-5


thus in the mixture [itex]\frac{PCl_3}{PCl_5}= \frac{25}{7} [/itex]

since 0.32258 mol of P reacted,


nPCl3 = 0.2520156 mol
nPCl5 = 0.070565 mol


so 39.365g of Cl2 was used.


Didn't check, so might have math error somewhere.
Hello!

Offline Rutherford

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Re: Practice problem 3
« Reply #2 on: April 24, 2014, 09:00:46 AM »
Well done ;D. That't correct. Now it's your turn to post a challenging problem.

Offline Big-Daddy

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Re: Practice problem 3
« Reply #3 on: April 26, 2014, 11:31:31 AM »
@Raderford : I'm a bit confused as to what you meant by "the consumption was 8.7 ml" and 7.1 ml respectively. You mean in your 10ml sample, adding 7.1 ml NaOH brought pH from initial to 3.5 and adding 8.7 ml NaOH brought pH from initial to 9.5?

(Btw - I think this cannot be the case - trying to solve the set of equations, taking equilibrium and everything, there is no solution for this set-up. But then SinkingTako appears to have done something different so maybe I am not understanding what the problem is asking?) Edit: Actually let me have another look...

4x + 6y = 0.0071 x 0.25
5x + 7y = 0.0087 x 0.25

What are the 4, 6, 5, 7 coefficients? How come your equations for determining number of moles of H3PO4 and H3PO3 have no dependence on the pH reached?
« Last Edit: April 26, 2014, 12:27:06 PM by Big-Daddy »

Offline Rutherford

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Re: Practice problem 3
« Reply #4 on: April 26, 2014, 12:31:09 PM »
Those are the pKa of indicators. From one mole of P(III) chloride you get 4 moles of acids and from one mole of P(V) chloride you get 6 moles of acids. Methyl orange changes color when HCl, H3PO4 and H3PO3 are neutralized (7.1 ml are spent). Phenolphthalein changes color when HPO42- and HPO32- are neutralized (8.7-7.1=1.6 ml to neutralize those two) spent. You don't need to consider equilibrium here, just simple acid-base titration stoichiometry.

Offline Big-Daddy

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Re: Practice problem 3
« Reply #5 on: April 26, 2014, 01:29:30 PM »
Yes makes sense on the whole, but I tried with equilibrium anyway and got a slightly different answer (I was wrong that it can't be solved, of course - I forgot to include HCl as an acidic source) (41.2 g to 3sf) without approximations. The fact it is so close makes me think it's probably right.
The difference could be down to incomplete neutralisation of H3PO4 at the end-point of methyl orange?

Offline Rutherford

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Re: Practice problem 3
« Reply #6 on: April 26, 2014, 01:45:54 PM »
Can;t say without seeing your work. Seems too much different.

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