# Question 5137a

Aug 22, 2015

The mole percent of potassium bromide in the mixture was $\text{34.8%}$.

#### Explanation:

FULL QUESTION

A mixture of sodium bromide and potassium bromide is quantitatively analysed by dissolving in water, adding dilute nitric acid and aqueous silver nitrate.

0.325 g of mixture gave 0.564 g of silver bromide. Calculate the mole percent of potassium bromide present in the mixture. (Na = 23, K = 39, Ag = 108, Br = 80)

The idea behind this problem is that the sodium bromide and the potassium bromide, both soluble compounds, will dissociate in aqueous solution to form sodium and potassium cations, and bromide anions.

These anions will react with the silver cations produced by the soluble silver nitrate to form the insoluble solid silver bromide, which precipitates out of solution.

Don't get distracted by the addition of the dilute nitric acid, it will have no effect on the ions that are of interest, i.e. ${\text{Ag}}^{+}$ and ${\text{Br}}^{-}$.

Since ${\text{Br}}^{-}$ is the conjugate base of a strong acid, it will not pick up a proton to form $\text{HBr}$. That means that yo ucan assume that all the bromide anions that are coming from the mixture will react with the silver cations coming from the silver nitrate.

The reactions that will interest you are

${\text{NaBr"_((aq)) -> "Na"_text((aq])^(+) + "Br}}_{\textrm{\left(a q\right]}}^{-}$

${\text{KBr"_((aq)) -> "K"_text((aq])^(+) + "Br}}_{\textrm{\left(a q\right]}}^{-}$

and

${\text{Ag"_((aq))^(+) + "Br"_text((aq])^(-) -> "AgBr}}_{\textrm{\left(s\right]}} \downarrow$

What you need to do is determine the percent composition of bromine in sodium bromide, potassium bromide, and silver bromide.

Using the molar masses given to you will get

(80color(red)(cancel(color(black)("g/mol"))))/((80 + 23)color(red)(cancel(color(black)("g/mol")))) * 100 = "77.7% Br" in $\text{NaBr}$

(80color(red)(cancel(color(black)("g/mol"))))/((80 + 39)color(red)(cancel(color(black)("g/mol")))) * 100 = "67.2% Br" in $\text{KBr}$

and

(80color(red)(cancel(color(black)("g/mol"))))/((80 + 108)color(red)(cancel(color(black)("g/mol")))) * 100 = "42.6% Br" in $\text{AgBr}$

Let's say that $x$ represents the mass of sodium bromide and $y$ the mass of potassium bromide. You know that you have

$x + y = \text{0.325 g}$

Moreover, you know that you collected 0.564 g of silver bromide, which means that you can determine how much bromine you had in solution

0.564color(red)(cancel(color(black)("g AgBr"))) * "42.6 g Br"/(100color(red)(cancel(color(black)("g AgBr")))) = "0.240 g Br"

This means that your second equation will be

$\frac{77.7}{100} \cdot x + \frac{67.2}{100} \cdot y = \text{0.240 g}$

Use the first equation to find $x$ as a function of $y$

$x = 0.325 - y$

The value of $y$ will be

$0.777 \cdot \left(0.325 - y\right) + 0.672 y = 0.240$

$0.253 - 0.777 y + 0.672 y = 0.240$

$- 0.105 y = - 0.013 \implies y = \text{0.124 g}$

This means that $x$ is equal to

$x = 0.325 - 0.124 = \text{0.201 g}$

Your mixture contained 0.201 g of sodium bromide and 0.124 g of potassium bromide. Use the two compounds' molar masses to get the number of moles of each

0.201color(red)(cancel(color(black)("g NaBr"))) * "1 mole"/(103color(red)(cancel(color(black)("g NaBr")))) = "0.00195 moles NaBr"

and

0.124color(red)(cancel(color(black)("g KBr"))) * "1 mole KBr"/(119color(red)(cancel(color(black)("g KBr")))) = "0.00104 moles KBr"

The total number of moles in the mixture was

${n}_{\text{total" = 0.00195 + 0.00104 = "0.00299 moles}}$

The mole percent of potassium bromide was

$\text{n/n%" = n_"KBr"/n_"total} \cdot 100$

(0.00104color(red)(cancel(color(black)("moles"))))/(0.00299color(red)(cancel(color(black)("moles")))) * 100 = color(green)("34.8%")#