# A solution is prepared by dissolving 0.5482 g of iron (III) nitrate in enough water to make a...

## Question:

A solution is prepared by dissolving 0.5482 g of iron (III) nitrate in enough water to make a 100.0 mL solution. A 10.00 mL aliquot (portion) of this solution is then diluted to a final volume of 250.0 mL. What is the final concentration of Fe{eq}^{3+} {/eq} ions in the final solution?

## Preparing Standard Solutions:

A common mistake students make when preparing standard solutions of a specific volume is to add the wrong amount of water. In our question, we are making a solution with a final volume of 100.0 mL. Some students incorrectly think this means adding 100.0 mL of water. In reality, the solute also has some volume. Thus the amount of water we actually add to give a total solution volume of 100.0 mL is less than 100.0 mL of water.

The question tells us we make a 100.0 mL solution from 0.5482 g of iron(III) nitrate and water. We first need to convert the mass of iron(III) nitrate to moles so that we can subsequently calculate the concentration of this original solution.

{eq}\rm n=\dfrac{m}{M}{/eq}

Where
n = moles
m = mass
M = molar mass

The molar mass of iron(III) nitrate ({eq}\rm Fe(NO_3)_3 {/eq}) is obtained by adding the atomic molar masses of each atom. The atomic molar masses are obtained from the periodic table:

{eq}\begin{align} \rm M &= \rm (55.85\:g/mol) + (3 \times 14.01\:g/mol) + (9 \times 16.00\:g/mol)\\ &= \rm 241.88\:g/mol \end{align} {/eq}

Now we can substitute the values in and solve:

{eq}\begin{align} \rm n &= \rm \dfrac{m}{M}\\ &= \rm \dfrac{0.5482\:g}{241.88\:g/mol}\\ &= \rm 0.002266\:mol \end{align} {/eq}

Now that we know the moles of solute, we can calculate the concentration of the solution:

{eq}\rm c=\dfrac{n}{V}{/eq}

Where
c = concentration
n = moles
V = volume

The volume must be in L, so we convert by:

{eq}\rm 100.0\:mL \times \dfrac{1\:L}{1000\:mL} = 0.1000\:L{/eq}

Now we can substitute in the values and solve:

{eq}\begin{align} \rm c&=\rm \dfrac{n}{V}\\ &= \rm \dfrac{0.002266\:mol}{0.1000\:L}\\ &= \rm 0.02266\:M \end{align} {/eq}

The question then tells us 10.00 mL of the above solution is diluted to a volume of 250.0 mL. We can use the following formula to find the final concentration of this solution:

{eq}\rm c_1V_1 = c_2V_2{/eq}

Where:

{eq}\rm c_1{/eq} is the initial concentration
{eq}\rm V_1{/eq} is the initial volume
{eq}\rm c_2{/eq} is the final concentration
{eq}\rm V_2{/eq} is the final volume

Rearranging the formula to solve for {eq}\rm c_2{/eq} and substituting in the values:

{eq}\begin{align} \rm c_2 =& \rm \dfrac{c_1V_1}{V_2}\\ &= \rm \dfrac{(0.02266\:M)(10.00\:mL)}{250\:mL}\\ &= \rm 9.066 \times 10^{-4}\:M \end{align} {/eq}

The above is the concentration of {eq}\rm Fe(NO_3)_3 {/eq}, which dissociates to form {eq}\rm Fe^{3+} {/eq} ions in a 1:1 ratio as per the equation below.

{eq}\rm Fe(NO_3)_3 \to Fe^{3+} + 3NO_3^- {/eq}

As a result the concentration we calculated for {eq}\rm Fe(NO_3)_3 {/eq} is equal to the {eq}\rm Fe^{3+} {/eq} concentration.

The {eq}\rm Fe^{3+} {/eq} ion concentration is {eq}\rm 9.066 \times 10^{-4}\:M {/eq}.

Calculating Dilution of Solutions

from

Chapter 8 / Lesson 5
69K

Learn what a solution is and how to properly dilute a new solution from a stock solution. Learn the dilution equation that combines molarity, the volume of stock solution and desired solution to determine how much stock solution is needed for the new solution.