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### Topic: Iodide solution  (Read 10615 times)

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#### Rutherford

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##### Iodide solution
« on: May 16, 2012, 10:02:23 AM »
In a iodide aqua solution of a monovalent metal a solution of AgNO3 is added until the separation of the AgI precipitate stopped. Mass of the obtained AgI is equal to the mass of the whole iodine solution. What was the mass share of AgNO3 in its solution?
How to calculate something if there aren't any numbers given?
I started with the equation:
xMI+xAgNO3 xAgI+xMNO3
(108+127)x=mr(of the iodide solution), but I don't have a clue what to do now.

#### Hunter2

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##### Re: Iodide solution
« Reply #1 on: May 17, 2012, 04:43:56 AM »
n1=m1/(x*M1) = m2/(y*M2) = n2

M is given by the silver salts and m1 = m2, x,y = 1 in this case

#### Rutherford

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##### Re: Iodide solution
« Reply #2 on: May 17, 2012, 06:51:49 AM »
"Mass of the obtained AgI is equal to the mass of the whole iodine solution."

#### Borek

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##### Re: Iodide solution
« Reply #3 on: May 17, 2012, 09:47:53 AM »
n1=m1/(x*M1) = m2/(y*M2) = n2

M is given by the silver salts and m1 = m2, x,y = 1 in this case

It always helps to explain symbols used, otherwise we are forced to make guesses. I suppose you mean m1 and m2 are masses of AgNO3 and AgI, if so, m1≠m2.

Perhaps I am missing something, but I don't see how to solve the problem. I can calculate concentration of iodide, but no idea how to calculate concentration of silver nitrate.
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#### whylolzz

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##### Re: Iodide solution
« Reply #4 on: May 17, 2012, 11:01:00 AM »
Hi, I tried attempting the question and reached an answer of some sort I think there is just enough material to work with to get an answer.
Do you have the actual answer you can provide so I can check whether it was right? Don't want to write it all out if it's wrong.

#### Rutherford

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##### Re: Iodide solution
« Reply #5 on: May 17, 2012, 11:46:25 AM »
Sadly I don't have , but it would be nice if you write how did you solve this.

#### whylolzz

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##### Re: Iodide solution
« Reply #6 on: May 17, 2012, 09:39:34 PM »
Actually, the question is quite vague. Is it stating that the mass of metal ['M' + AgNO3] = [Mass of iodine solution]?

#### Borek

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##### Re: Iodide solution
« Reply #7 on: May 18, 2012, 02:38:55 AM »
The way I read it it states "mass of the precipitated AgI = initial mass of the iodide solution".
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#### Rutherford

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##### Re: Iodide solution
« Reply #8 on: May 18, 2012, 02:41:48 AM »
Yes, it is so.

#### whylolzz

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##### Re: Iodide solution
« Reply #9 on: May 18, 2012, 03:02:12 AM »
Yes, but when it says iodide solution, is the iodide solution just [MI] or [MI + AgNO3].

#### Borek

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##### Re: Iodide solution
« Reply #10 on: May 18, 2012, 03:23:54 AM »
MI IMHO.
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#### whylolzz

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##### Re: Iodide solution
« Reply #11 on: May 18, 2012, 04:18:08 AM »
$$Let\;mass\;of\;AgNO_{3}=x \\ \\ Let\;mass\;of\;MI=AgI=y \\ \\ n_{AgNO_{3}}=\frac{x}{169.88} \\ \\ AgNO_{3}\;\;is\;the\;limiting\;reagent\;as\;the\;question\;implies \\ \\ n_{AgI}=\frac{y}{234.77} \\ \\ \;\;But\;n_{AgI}=n_{AgNO_{3}} \\ \\ \therefore\;\frac{x}{169.88}=\frac{y}{234.77} \\ \\ \frac{x}{y}=0.72 \\ \\ \;hence\;72\%\;of\;the\;solution\;is\;AgNO_{3}$$

this is what I get as I interpret the question. However, I'm not sure whether the mass share is including AgNO3
« Last Edit: May 18, 2012, 04:34:48 AM by whylolzz »

#### Borek

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##### Re: Iodide solution
« Reply #12 on: May 18, 2012, 05:46:55 AM »
Looks to me like you are mistaking mass of the solution with mass of the solute. Yes, mass of the AgNO3 is 72% of the mass of AgI that can be produced from that AgNO3, but it has nothing to do with the question.

Note: don't use LaTeX for whole post, use it just for formulas (and not for chemical ones, unless really necessary). It looks better that way.

Radeford, can you post the original question without translating it to English?
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#### Rutherford

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##### Re: Iodide solution
« Reply #13 on: May 18, 2012, 05:59:20 AM »
How would you understand?

#### Borek

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##### Re: Iodide solution
« Reply #14 on: May 18, 2012, 06:43:40 AM »
How would you understand?

Don't worry about that - if I will not understand it won't push you back, if I will understand, it may help. You have nothing to lose.
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