Chapter 4, Problem 161MP

You have two 500.0-mL aqueous solutions. Solution A is a solution of a metal nitrate that is 8.246% nitrogen by mass. The ionic compound in solution B consists of potassium, chromium, and oxygen; chromium has an oxidation state of + 6 and there are 2 potassiums and 1 chromium in the formula. The masses of the solutes in each of the solutions are the same. When the solutions are added together, a blood-red precipitate forms. After the reaction bas gone to completion, you dry the solid and find that it has a mass of 331.8 g.

a. Identify the ionic compounds in solution A and solution B.

b. Identify the blood-red precipitate.

c. Calculate the concentration (molarity) of all ions in the original solutions.

d. Calculate the concentration (molarity) of all ions in the final solution.

Interpretation Introduction

Interpretation: The ionic compounds in solution A and B has to be identified.

Concept Introduction: When two soluble solutions are mixed together, an insoluble salt formation occurs called as precipitate. These precipitates fall out of the solution and the reactions are called as precipitation reaction.

Concentration of solution can be defined in terms of molarity as moles of solute (in grams) to the volume of solution (in litres). The concentration of solution can be calculated by,

$\text{Concentration(M)=}\frac{\text{Moles}\text{of}\text{solute(g)}}{\text{Volume}\text{of}\text{solution(L)}}$

Answer to Problem 161MP

The ionic compound in solution A is ${\text{AgNO}}_{\text{3}}$ and ionic compound B is ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$.

Explanation of Solution

Given:

Record the given data

Volume of two aqueous solution   = $\text{500}\text{.0}\text{mL}$

Mass percent of Nitrogen in solution A = $\text{8}\text{.246\%}$

Mass of precipitate   = $\text{331}\text{.8}\text{g}$

The volume of two aqueous solutions along with mass percent of Nitrogen in solution A of metal nitrate and mass of precipitate are recorded as shown above.

To identify the ionic compound in solution A

Molecular mass of oxygen = $\text{16}\text{.00}\text{g}$

Molecular mass of nitrogen = $\text{14}\text{.01}\text{g}$

Compound A contains metal nitrate ${\text{M(NO}}_{\text{3}}{\text{)}}_{\text{x}}$

Assuming the mass of metal nitrate is ${\text{M(NO}}_{\text{3}}{\text{)}}_{\text{x}}$= $\text{100}\text{.0}\text{g}$

The moles of oxygen is calculated using

$\begin{array}{l}\text{Moles}\text{of}\text{oxygen=8}\text{.246}\text{g}\text{N×}\frac{\text{48}\text{.00}\text{g}\text{O}}{\text{14}\text{.01}\text{g}\text{N}}\\ \text{=}\text{28}\text{.25}\text{g}\text{O}\\ \end{array}$

Thus, the mass of nitrate = $\text{8}\text{.246+28}\text{.25}\text{g=36}\text{.50}\text{g(if}\text{x=1)}$

Case 1:

$\begin{array}{l}\text{x=1:}\text{mass}\text{of}\text{M=100}\text{.00-36}\text{.50}\text{g}\\ \text{=63}\text{.50}\text{g}\end{array}$

$\text{Mol}\text{M=mol}\text{N=}\frac{\text{8}\text{.246}\text{g}}{\text{14}\text{.01}\text{g/mol}}\text{=0}\text{.5886}\text{mol}$

$\begin{array}{l}\text{Molar}\text{mass}\text{of}\text{metal}\text{M=}\frac{\text{63}\text{.50}\text{g}}{\text{0}\text{.5886}\text{mol}}\\ \text{=107}\text{.9}\text{g/mol}\end{array}$

The metal M is identified as Silver $\text{(Ag)}$

Case 2:

$\begin{array}{l}\text{x=2}\text{:}\text{Mass}\text{of}\text{M=100}\text{.00-2(36}\text{.50)}\\ \text{=27}\text{.00}\text{g}\end{array}$

$\text{Mol}\text{M=}\frac{\text{1}}{\text{2}}\text{mol}\text{N=}\frac{\text{0}\text{.5886}\text{mol}}{\text{2}}\text{=0}\text{.2943}\text{mol}$

$\begin{array}{l}\text{Molar}\text{mass}\text{of}\text{metal}\text{M=}\frac{\text{27}\text{.00}\text{g}}{\text{0}\text{.2943}\text{mol}}\\ \text{=}\text{91}\text{.74}\text{g/mol}\end{array}$

This is close to $\text{Zr}$ but $\text{Zr}$ doesn’t form stable $\text{+2}$ ions in the solution, and therefore it forms stable $\text{+4}$ ions.

x would not take the value of 3, since the nitrates present would have a higher mass than $\text{100}\text{.0}\text{g}$ .

Therefore, the compound A is identified as ${\text{AgNO}}_{\text{3}}$

The ionic compound is solution A is calculated by plugging in the values of mass of Nitrogen in metal nitrate and also by calculating the mass of M and its molar mass. The value of calculated molar mass is matched with value of closest molar mass of element in the periodic table. The ionic compound is identified as ${\text{AgNO}}_{\text{3}}$.

To identify the compound B

Oxidation state of chromium = $\text{+6}$

The compound B has two Potassium atoms and one Chromium atom. Hence, the formula can be given as, ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{x}}$ .

The oxidation state of oxygen is calculated by,

$\begin{array}{l}\text{6+x(-2)=-2}\\ \text{x=4}\end{array}$

Therefore, the formula becomes ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$

Compound B is identified ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$

The ionic compound in solution B is identified by calculating the oxidation state of Oxygen and substituting in the unknown x in the formula. The ionic compound in solution B is identified as ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$.

Interpretation Introduction

Interpretation: The blood-red precipitate has to be identified.

Concept Introduction: When two soluble solutions are mixed together, an insoluble salt formation occurs called as precipitate. These precipitates fall out of the solution and the reactions are called as precipitation reaction.

Concentration of solution can be defined in terms of molarity as moles of solute (in grams) to the volume of solution (in litres). The concentration of solution can be calculated by,

$\text{Concentration(M)=}\frac{\text{Moles}\text{of}\text{solute(g)}}{\text{Volume}\text{of}\text{solution(L)}}$

Answer to Problem 161MP

The blood red precipitate is ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$.

Explanation of Solution

To identify the blood red precipitate

The blood red precipitate is identified as Silver chromate. A precipitate of silver chromate is formed when aqueous solution of Silver nitrate is reacted with aqueous solution of Potassium chromate. Potassium nitrate remains as spectator ions inside the solution.

The chemical equation can be given as,

${\text{2AgNO}}_{\text{3(aq)}}{\text{+K}}_{\text{2}}{\text{CrO}}_{\text{4(aq)}}\to {\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4(s)}}{\text{+2KNO}}_{\text{3(aq)}}$

Interpretation Introduction

Interpretation: The concentration of original solution has to be calculated.

Concept Introduction: When two soluble solutions are mixed together, an insoluble salt formation occurs called as precipitate. These precipitates fall out of the solution and the reactions are called as precipitation reaction.

Concentration of solution can be defined in terms of molarity as moles of solute (in grams) to the volume of solution (in litres). The concentration of solution can be calculated by,

$\text{Concentration(M)=}\frac{\text{Moles}\text{of}\text{solute(g)}}{\text{Volume}\text{of}\text{solution(L)}}$

Answer to Problem 161MP

The concentration of all ions in original solution are

$\begin{array}{l}\text{Solution}{\text{A,Ag}}^{\text{+}}\text{=4}\text{.000}\text{M}{\text{Ag}}^{\text{+}}\\ {\text{NO}}_{\text{3}}{}^{\text{-}}\text{=4}\text{.000}\text{M}{\text{NO}}_{\text{3}}{}^{\text{-}}\\ \text{Solution}\text{B,}{\text{K}}^{\text{+}}\text{=7}\text{.000}\text{M}{\text{K}}^{\text{+}}\\ {\text{CrO}}_{\text{4}}{}^{\text{2-}}\text{=3}\text{.500}\text{M}{\text{CrO}}_{\text{4}}{}^{\text{2-}}\end{array}$

Explanation of Solution

Calculate the concentration of ions in original solutions.

Mass of precipitate = $\text{331}\text{.8}\text{g}$

Molar mass of ${\text{AgNO}}_{\text{3}}\text{=169}\text{.9}\text{g}$

Molar mass of ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{=194}\text{.2}\text{g}$

The chemical equation can be given as,

${\text{2AgNO}}_{\text{3(aq)}}{\text{+K}}_{\text{2}}{\text{CrO}}_{\text{4(aq)}}\to {\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4(s)}}{\text{+2KNO}}_{\text{3(aq)}}$

From the chemical equation, 2 moles of ${\text{AgNO}}_{\text{3}}$ and 1 mole of ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ reacts to yield 1 mole ( $\text{331}\text{.8}\text{g}$) of ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4(s)}}$.1 mole of precipitate is formed since $\text{331}\text{.8}\text{g}$ of ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4(s)}}$ is produced that is equal to molar mass of ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4(s)}}$.

$\begin{array}{l}\text{Mass}\text{of}{\text{AgNO}}_{\text{3}}\text{=2}\text{.000}\text{mol}{\text{AgNO}}_{\text{3}}\text{×}\frac{\text{169}\text{.9}\text{g}}{\text{mol}}\text{=339}\text{.8}\text{g}{\text{AgNO}}_{\text{3}}\\ \\ \text{Mass of}{\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{=1}\text{.000}\text{mol}{\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{×}\frac{\text{194}\text{.2}\text{g}}{\text{mol}}\text{=194}\text{.2}\text{g}{\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}\end{array}$

From the given question, the reactants are said to have equal masses.

Therefore, the sum of reactant masses must be either,

$\text{339}\text{.8}\text{g}{\text{AgNO}}_{\text{3}}\text{+339}\text{.8}\text{g}{\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ or  $\text{194}\text{.2}\text{g}{\text{AgNO}}_{\text{3}}\text{+194}\text{.2}\text{g}{\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$

Assuming that, $\text{194}\text{.2}\text{g}$ quantities are correct

Then, when $\text{194}\text{.2}\text{g}{\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{(1}\text{mol)}\text{reacts}\text{with}\text{339}\text{.8}\text{g}{\text{AgNO}}_{\text{3}}\text{(2}\text{mol)}$ must be present to react with all ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$.

But we assumed only 194.2 g AgNO3 is present; this cannot be correct. Instead of ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ limiting, ${\text{AgNO}}_{\text{3}}$ must be limiting, and we reacted 339.8 g ${\text{AgNO}}_{\text{3}}$ and 339.8 g ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$

Therefore, the concentration of ions in original solution,

$\begin{array}{l}\text{Solution}\text{A,}\\ {\text{Molarity of Ag}}^{\text{+}}\text{=}\frac{\text{2}\text{.000}\text{mol}{\text{Ag}}^{\text{+}}}{\text{0}\text{.5000}\text{L}}\text{=4}\text{.000}\text{M}{\text{Ag}}^{\text{+}}\\ \text{Molarity}\text{of}{\text{NO}}_{\text{3}}{}^{\text{-}}\text{=}\frac{\text{2}\text{.000}\text{mol}{\text{NO}}_{\text{3}}{}^{\text{-}}}{\text{0}\text{.5000}\text{L}}\text{=4}\text{.000}\text{M}{\text{NO}}_{\text{3}}{}^{\text{-}}\\ \\ \text{Solution}\text{B,}\\ {\text{Moles of K}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{=339}\text{.8}\text{g}{\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{×}\frac{\text{1}\text{mol}}{\text{194}\text{.2}\text{g}}\text{=1}\text{.750}\text{mol}{\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}\\ \text{Molarity}\text{of}{\text{K}}^{\text{+}}\text{=}\frac{\text{2×1}\text{.750}{\text{molK}}^{\text{+}}}{\text{0}\text{.5000}\text{L}}\text{=7}\text{.000}\text{M}{\text{K}}^{\text{+}}\\ \\ \text{Molarity}\text{of}{\text{CrO}}_{\text{4}}{}^{\text{2-}}\text{=}\frac{\text{1}\text{.750}\text{mol}{\text{CrO}}_{\text{4}}{}^{\text{2-}}}{\text{0}\text{.5000}\text{L}}\text{=3}\text{.500}\text{M}{\text{CrO}}_{\text{4}}{}^{\text{2-}}\end{array}$

The concentrations of individual ions are calculated by plugging in the values of moles taking part in the chemical reaction to the volumes of solutions. The concentrations of individual ions in original solution are found to be $\text{4}\text{.000}\text{M}{\text{Ag}}^{\text{+}}$, $\text{4}\text{.000}\text{M}{\text{NO}}_{\text{3}}{}^{\text{-}}$, $\text{7}\text{.000}\text{M}{\text{K}}^{\text{+}}$ and $\text{3}\text{.500}\text{M}{\text{CrO}}_{\text{4}}{}^{\text{2-}}$

Interpretation Introduction

Interpretation: The concentration of final solution has to be calculated.

Concept Introduction: When two soluble solutions are mixed together, an insoluble salt formation occurs called as precipitate. These precipitates fall out of the solution and the reactions are called as precipitation reaction.

Concentration of solution can be defined in terms of molarity as moles of solute (in grams) to the volume of solution (in litres). The concentration of solution can be calculated by,

$\text{Concentration(M)=}\frac{\text{Moles}\text{of}\text{solute(g)}}{\text{Volume}\text{of}\text{solution(L)}}$

Answer to Problem 161MP

The concentration of all ions in final solution are

                         $\begin{array}{l}{\text{M}}_{{\text{K}}^{\text{+}}}\text{=3}\text{.500}\text{M}{\text{K}}^{\text{+}}{\text{M}}_{{\text{NO}}_{\text{3}}{}^{\text{-}}}\text{=2}\text{.000}\text{M}{\text{NO}}_{\text{3}}{}^{\text{-}}\\ {\text{M}}_{{\text{CrO}}_{\text{4}}{}^{\text{2-}}}\text{=0}\text{.750}\text{M}{\text{CrO}}_{\text{4}}{}^{\text{2-}}{\text{M}}_{{\text{Ag}}^{\text{+}}}\text{=0}\text{M(limiting}\text{reagent)}\end{array}$

Explanation of Solution

To calculate the concentration of ions in final solution

The moles of spectator ions ( ${\text{K}}^{\text{+}}\text{and}{\text{NO}}_{\text{3}}{}^{\text{-}}$) remains unchanged after the reaction.

${\text{Ag}}^{\text{+}}$ is limiting reagent and hence the molarity of ${\text{Ag}}^{\text{+}}$ will be zero after the precipitation is complete.

$\begin{array}{l}{\text{Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}\text{+}{\text{CrO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4(s)}}\\ \text{Initial}\text{2}\text{.000}\text{mol}\text{1}\text{.750}\text{mol}\text{0}\\ \text{Change}\text{-2}\text{.000}\text{mol}\text{-1}\text{.000}\text{mol}\text{+1}\text{.000}\text{mol}\\ \text{After}\\ \text{reaction}\text{0}\text{0}\text{.750}\text{mol}\text{1}\text{.000}\text{mol}\end{array}$

Therefore, the concentrations of individual ions in final solution are

$\begin{array}{l}\text{Molarity}\text{of}{\text{K}}^{\text{+}}\text{=}\frac{\text{2×1}\text{.750}\text{mol}}{\text{1}\text{.0000}\text{L}}\text{=3}\text{.500}\text{M}{\text{K}}^{\text{+}}\\ \text{Molarity}\text{of}{\text{NO}}_{\text{3}}{}^{\text{-}}\text{=}\frac{\text{2}\text{.000}\text{mol}}{\text{1}\text{.0000}\text{L}}\text{=2}\text{.000}\text{M}{\text{NO}}_{\text{3}}{}^{\text{-}}\\ \text{Molarity}\text{of}{\text{CrO}}_{\text{4}}{}^{\text{2-}}\text{=}\frac{\text{0}\text{.750}\text{mol}}{\text{1}\text{.0000}\text{L}}\text{=0}\text{.750}\text{M}{\text{CrO}}_{\text{4}}{}^{\text{2-}}\\ \text{Molarity}\text{of}{\text{Ag}}^{\text{+}}\text{=}\text{0M(limiting}\text{reagent)}\end{array}$

The concentrations of individual ions in final solution are calculated by plugging in values of their respective moles to the volume of the solution. The molarity of individual ions in final solution are found to be $\text{3}\text{.500}\text{M}{\text{K}}^{\text{+}}$, $\text{2}\text{.000}\text{M}{\text{NO}}_{\text{3}}{}^{\text{-}}$, $\text{0}\text{.750}\text{M}{\text{CrO}}_{\text{4}}{}^{\text{2-}}$ and ${\text{0M Ag}}^{\text{+}}$ respectively.