Question a373a

Apr 27, 2017

${\text{0.05 mg AgNO}}_{3}$

Explanation:

We use parts per million to describe the concentration of a solution that contains very small amounts of solute, i.e. trace amounts.

The general idea behind using parts per million is that we look at the number of parts of solute present for every

${10}^{6} = 1 , 000 , 000$

parts of solution. In your case, a $\text{0.1 ppm}$ silver nitrate solution would contain $0.1$ parts of solute for every ${10}^{6}$ parts of solution.

Now, you can assume that the density of this solution is equal to the density of water, which you can approximate as ${\text{1 g mL}}^{- 1}$. This means that your solution must contain

$\text{0.1 g AgNO"_3 -> "0.1 parts of solute}$

for every

${10}^{6} \textcolor{w h i t e}{.} \text{g solution}$

The mass of the solution will be equal to

300 color(red)(cancel(color(black)("mL"))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = "300 g"

At this point, you can use the solution's ppm concentration as a conversion factor to determine how many grams of silver(I) cations it must contain

300 color(red)(cancel(color(black)("g solution"))) * overbrace("0.1 g Ag"^(+)/(10^6color(red)(cancel(color(black)("g solution")))))^(color(blue)("= 0.1 ppm solution")) = 3 * 10^(-5) ${\text{g Ag}}^{+}$

This is equivalent to

3 * 10^(-5) color(red)(cancel(color(black)("g"))) * (10^3 color(white)(.)"mg")/(1color(red)(cancel(color(black)("g")))) = "0.03 mg Ag"^(+)

Now, in order to figure out how much silver nitrate will produce this much silver(I) cations in solution, use the compound's percent composition.

Silver nitrate has a molar mass of ${\text{169.87 g mol}}^{- 1}$ and elemental silver, $\text{Ag}$, has a molar mass of ${\text{107.87 g mol}}^{- 1}$.

Since every $1$ mole of silver nitrate contains $1$ mole of silver, you can say that the percent composition of silver in this compound is

(107.87 color(red)(cancel(color(black)("g mol"^(-1)))))/(169.87color(red)(cancel(color(black)("g mol"^(-1))))) * 100% = "63.5% Ag"#

This means that every $\text{100 g}$ of silver nitrate contain $\text{63.5 g}$ of silver, which is equivalent to saying that for every $\text{100 mg}$ of silver nitrate you get $\text{63.5 mg}$ of silver.

You can say that the mass of the silver(I) cations is approximately equal to the mass of elemental silver, which means that in order to get $\text{0.03 mg}$ of silver(I) cations in your solution, you must dissolve

$0.03 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{mg Ag"^(+)))) * "100 mg AgNO"_3/(63.5 color(red)(cancel(color(black)("mg Ag"^(+))))) = color(darkgreen)(ul(color(black)("0.05 mg AgNO}}_{3}}}}$

The answer is rounded to one significant figure.

Therefore, you can say that a solution that contains $\text{0.05 mg}$ of silver nitrate in $\text{300 mL}$ of solution will have a concentration of $\text{0.1 ppm}$ of silver(I) cations, ${\text{Ag}}^{+}$.