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Topic: Oxide+water=unknown acid  (Read 2036 times)

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Offline Rutherford

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Oxide+water=unknown acid
« on: February 23, 2013, 05:53:14 AM »
1g of oxide A reacted with water to form 1.49g of acid B. The oxide reacted completely. The formed acid is a solid substance. For the neutralization of 0.149g of acid B, 36.34ml of NaOH (c=0.1M) were spent. Write the formulas of the acid B and oxide A.

Using the data from the titration, the equivalent mass of the acid is 41g/mol. If it is a diprotic acid, M=82g/mol. With a little guessing, I got H2PHO3, which is correct.
I didn't use the masses of the oxide and acid, I just solved this on guessing. How to solve this by using the masses data (not by guessing)?
xA + yH2O :rarrow: B
1g  0.49g   1.49g
What to do now?

Offline Borek

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Re: Oxide+water=unknown acid
« Reply #1 on: February 23, 2013, 06:43:06 AM »
This type of problems always require a little bit of guessing.

xA + yH2O :rarrow: B
1g  0.49g   1.49g

There are infinitely many pairs of x & y that fit the equation (to some extent it can be treated as a diophantine equation), luckily only some have chemical sense.
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Offline Rutherford

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Re: Oxide+water=unknown acid
« Reply #2 on: February 23, 2013, 07:23:35 AM »
I don't understand what you mean. Which one will have sense?

Offline Borek

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Re: Oxide+water=unknown acid
« Reply #3 on: February 23, 2013, 08:39:45 AM »
Those for which exist compounds with correct molar masses (and reacting in the given ratio).
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Offline Rutherford

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Re: Oxide+water=unknown acid
« Reply #4 on: February 23, 2013, 08:57:38 AM »
Then it seems better to just guess with the molar mass of 82g/mol. If the compound has 3 O then the mass that left is 34g/mol, P+3H.

Never mind then and thanks.

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