Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Rutherford on May 16, 2012, 10:02:23 AM
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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 :rarrow: xAgI+xMNO3
(108+127)x=mr(of the iodide solution), but I don't have a clue what to do now.
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n1=m1/(x*M1) = m2/(y*M2) = n2
M is given by the silver salts and m1 = m2, x,y = 1 in this case
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"Mass of the obtained AgI is equal to the mass of the whole iodine solution."
I don't understand your way.
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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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Hi, I tried attempting the question and reached an answer of some sort :P 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.
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Sadly I don't have :( , but it would be nice if you write how did you solve this.
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Actually, the question is quite vague. Is it stating that the mass of metal ['M' + AgNO3] = [Mass of iodine solution]?
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The way I read it it states "mass of the precipitated AgI = initial mass of the iodide solution".
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Yes, it is so.
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Yes, but when it says iodide solution, is the iodide solution just [MI] or [MI + AgNO3].
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MI IMHO.
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$$ Let\;mass\;of\;AgNO_{3}=x
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Let\;mass\;of\;MI=AgI=y
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n_{AgNO_{3}}=\frac{x}{169.88}
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AgNO_{3}\;\;is\;the\;limiting\;reagent\;as\;the\;question\;implies
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n_{AgI}=\frac{y}{234.77}
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\;\;But\;n_{AgI}=n_{AgNO_{3}}
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\therefore\;\frac{x}{169.88}=\frac{y}{234.77}
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\frac{x}{y}=0.72
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\;hence\;72\%\;of\;the\;solution\;is\;AgNO_{3} $$
this is what I get as I interpret the question. However, I'm not sure whether the mass share is including AgNO3
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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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How would you understand?
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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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OK, I got the question in PM. I did understood the question correctly:
Silver nitrate solution was added to the solution of monovalent metal iodide till the precipitation stopped. Mass of the produced silver iodide precipitate was identical to the mass of the iodide solution. What was the mass fraction of silver nitrate in its solution?
That's the question I tried to solve, and I don't see how it can be done. It is possible to calculate concentration of the iodide solution, although it requires assuming its density is 1 g/mL. This is unrealistic, as it yields iodide concentration in the range of 50%, sodium and potassium iodides of this concentration have density around 1.5 g/mL.
I wonder why you didn't want to post the original question on the forum - is it because you are not allowed to seek help?
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Thanks anyway for the try. Didn't want to post because of private policy.
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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 :rarrow: xAgI+xMNO3
(108+127)x=mr(of the iodide solution), but I don't have a clue what to do now.
This is a multisolution problem. Take for example 1 mole AgI as a resulting mass. From the stoichiometry of reaction a solution of iodide should contain 1 mole of (LiI, NaI, KI, RbI or ewentually NH4I) and enough of water to the whole mass 108+127=235 g.
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Then 1 mole of AgI is made from 170g of AgNO3. M+127+m(H2O)=235g, but I need the mass of water in the AgNO3 solution.