Chapter 21, Problem 21.201QP

(a)

Interpretation Introduction

Interpretation:

The complete and balanced equation for given combustion reaction has to be written.

Concept introduction:

Balanced Chemical equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

Answer to Problem 21.201QP

${\text{P}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{ + 5O}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{}_{\left(\text{s}\right)}$

Explanation of Solution

To write: The complete and balanced equation for given combustion.

${\text{P}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{ + 5O}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{}_{\left(\text{s}\right)}$

When white phosphorous was heated in presence of more oxygen it gives phosphorous pentoxide.

(b)

Interpretation Introduction

Interpretation:

The amount of white phosphorous obtained from combustion reaction has to be calculated.

Concept introduction:

• Acid dissociation constant ${\text{K}}_{\text{a}}$ is defined as measure of acid strength in a given solution
• ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[A}}^{\text{-}}{\text{][H}}^{\text{+}}\text{]}}{\text{[HA]}}$
• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
•   $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

Answer to Problem 21.201QP

The amount of white phosphorous obtained from combustion reaction is $\text{0}\text{.69 g P}$.

Explanation of Solution

To calculate: The amount of white phosphorous obtained from combustion reaction.

$\begin{array}{l}{\text{P}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{ + 5O}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{}_{\left(\text{s}\right)}\\ \\ {\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{ + H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\rightleftarrows }{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{ + H}}_{\text{2}}{\text{PO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\text{,}{\text{K}}_{\text{a}}\text{= 6}\text{.9}\times {\text{10}}^{\text{-3}}\end{array}$

When white phosphorous was heated in presence of more oxygen it gives phosphorous pentoxide.  ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ gives phosphoric acid when P₄O₁₀ dissolves in water, which undergoes dissociation and forms acid solution.

The equilibrium constant expression for above reaction is,

$\begin{array}{l}{\text{K}}_{\text{a}}\text{= }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}\text{]}}{{\text{[H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{]}}\\ \\ \text{=}\frac{{\text{(x)}}^{\text{2}}}{{{\text{[H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{]}}_{\text{o}}\text{- x}}\end{array}$

By using the pH, the x value is calculated as

$\text{x= }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{ = 10}}^{\text{-pH}}{\text{= 10}}^{\text{-2}\text{.091}}\text{= 8}\text{.109}\text{×}{\text{10}}^{\text{-3}}\text{M}$

By rearranging the equilibrium constant expression we get phosphoric acid concentration.

$\begin{array}{l}{\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]}_{\text{o}}\text{=}\text{x}\text{+ }\frac{{\text{(x)}}^{\text{2}}}{{\text{K}}_{\text{a}}}\\ \\ \text{= 8}\text{.109}\text{×}{\text{10}}^{\text{-5}}\text{M+}\frac{{\text{(1}{\text{.071 × 10}}^{\text{-5}}\text{)}}^{\text{2}}}{\text{5}{\text{.882 × 10}}^{\text{-10}}}\\ \\ \text{= 0}\text{.01764M}\end{array}$

The amount of phosphorous present in given sample is

$\left(\text{0}\text{.01764M}\right)\text{×}\left(\text{1}\text{.258 L}\right)\text{×}\frac{\text{1 mol P}}{{\text{1 mol H}}_{\text{3}}{\text{PO}}_{\text{4}}}\text{×}\frac{\text{30}\text{.97 g}}{\text{1 mol P}}\text{= 0}\text{.69 g P}$

(c)

Interpretation Introduction

Interpretation:

The complete and balanced equation for reaction between phosphoric acid and calcium nitrate has to be written.

Concept introduction:

Balanced Chemical equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

Answer to Problem 21.201QP

${\text{2H}}_{\text{3}}{\text{PO}}_{\text{4}}{}_{\left(\text{aq}\right)}\text{ + 3Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}{}_{\left(\text{s}\right)}{\text{ + 6HNO}}_{\text{3}}{}_{\left(\text{aq}\right)}$

Explanation of Solution

To write: The complete and balanced equation for reaction between phosphoric acid and calcium nitrate.

${\text{2H}}_{\text{3}}{\text{PO}}_{\text{4}}{}_{\left(\text{aq}\right)}\text{ + 3Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}{}_{\left(\text{s}\right)}{\text{ + 6HNO}}_{\text{3}}{}_{\left(\text{aq}\right)}$

When phosphoric acid reacts with calcium nitrate it forms calcium phosphate and nitric acid.

(d)

Interpretation Introduction

Interpretation:

The amount of calcium phosphate formed when phosphoric acid reacts with calcium nitrate has to be calculated.

Answer to Problem 21.201QP

The amount of calcium phosphate obtained is $\text{3}\text{.4 g}$.

Explanation of Solution

To calculate: The amount of calcium phosphate formed when phosphoric acid reacts with calcium nitrate.

${\text{2H}}_{\text{3}}{\text{PO}}_{\text{4}}{}_{\left(\text{aq}\right)}\text{ + 3Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}{}_{\left(\text{s}\right)}{\text{ + 6HNO}}_{\text{3}}{}_{\left(\text{aq}\right)}$

When phosphoric acid reacts with calcium nitrate it forms calcium phosphate and nitric acid.

The amount of calcium phosphate is calculated using the stoichiometry of reactants and products.

$\left(\text{0}\text{.01764M}\right)\text{×}\left(\text{1}\text{.258 L}\right)\text{×}\frac{\text{1 mol P}}{{\text{1 mol H}}_{\text{3}}{\text{PO}}_{\text{4}}}\text{×}\frac{\text{310}\text{.18 g}}{{\text{1 mol Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}}\text{=}\text{3}\text{.4}\text{g}$