Chapter 19, Problem 88QRT

(a)

Interpretation Introduction

Interpretation:

For burning of phosphorus in excess oxygen, balanced chemical equation has to be written.

Explanation of Solution

Phosphorus belongs to Group VA in periodic table.  It is a non-metal.  Non-metals burns with oxygen to attain highest possible oxidation state.  Highest oxidation state of phosphorus is $+5$.

When phosphorus burns in oxygen, phosphorus pentoxide is formed.  This can be represented as,

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

Balanced chemical equation can be given as,

    $\begin{array}{l}{\text{P}}_{\text{4}}(s)+5{\text{O}}_{\text{2}}(g)\to 2{\text{P}}_{\text{2}}{\text{O}}_{\text{5}}(s)(or)\\ {\text{P}}_{\text{4}}(s)+5{\text{O}}_{\text{2}}(g)\to {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}(s)\end{array}$

(b)

Interpretation Introduction

Interpretation:

$\text{pH}$ of the $\text{250}\text{mL}$ solution, that is formed when phosphorus pentoxide reacts with water has to be calculated.

Explanation of Solution

Non-metals burns in excess oxygen to form oxides that are acidic in nature.  This means that they can react with water to form aqueous acid.  Phosphorus pentoxide reacts with water to form phosphoric acid.  This can be represented as,

    ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}(s)+{\text{6H}}_{\text{2}}\text{O}(l)\to 4{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)$

Molarity of phosphoric acid formed when $\text{5}\text{.00}\text{g}$ of white phosphorus is taken as starting material can be calculated as shown below,

    $\begin{array}{c}\frac{\text{5}\text{.00}\text{g}{\text{P}}_{\text{4}}\times \frac{\text{1}\text{mol}{\text{P}}_{\text{4}}}{\text{123}\text{.8952}\text{g}{\text{P}}_{\text{4}}}\times \frac{\text{1}\text{mol}{\text{P}}_{\text{4}}{\text{O}}_{\text{10}}}{\text{1}\text{mol}{\text{P}}_{\text{4}}}\times \frac{\text{4}\text{mol}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}{\text{1}\text{mol}{\text{P}}_{\text{4}}{\text{O}}_{\text{10}}}}{0.250\text{L}}\text{=}\frac{\frac{\text{20}}{\text{123}\text{.8952}}\text{mol}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}{\text{0}\text{.250}\text{L}}\\ \text{=}\frac{\text{0}\text{.1614}}{\text{0}\text{.25}}\text{M}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\\ =0.645\text{6}\text{M}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\end{array}$

$\text{pH}$ of phosphoric acid solution will be affected by the first ionization only as phosphoric acid is a weak acid.  This equilibrium reaction can be represented as,

    ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)+{\text{H}}_{\text{2}}\text{O}(l)\to {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}+{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}(aq)$

Equilibrium expression can be represented for the above reaction as,

    $K_{a1}=\frac{[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}][{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}]}{[\text{H3PO4}]}$

The concentration of products increases stoichiometrically until the equilibrium is reached.

${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$         ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}$        ${\text{N}}_{\text{3}}{}^{-}\text{(aq)}$

Initial        $\text{0}\text{.6456}\text{M}$        $0$            $\text{0}$

Change    $-x$            $+x\text{M}$            $+x\text{M}$

Equilibrium    $\text{0}\text{.6456}\text{M}-x$        $x\text{M}$            $x\text{M}$

Equilibrium constant for phosphoric acid is $7.2\times {10}^{-3}$.  Therefore, at equilibrium,

    $\begin{array}{c}{K}_{a1}=\frac{(x)(x)}{(0.010-x)}\\ 7.2\times {10}^{-3}=\frac{x^2}{0.010-x}\\ x^2=(7.2\times {10}^{-3})(0.010-x)\end{array}$

Solving the above equation,

    $\begin{array}{c}{x}^2+(7.2\times {10}^{-3})x-(7.2\times {10}^{-5})=0\\ x=0.0056\text{M}\end{array}$

Value of $x$ is the concentration of the hydronium ion.  This means,

    ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]=0}\text{.0056}\text{M}$

$\text{pH}$ can be calculated by taking negative logarithm for the concentration of hydronium ion.  This can be shown as,

    $\begin{array}{c}\text{pH}=-\log [{\text{H}}_{\text{3}}{\text{O}}^{+}]\\ =-\log [0.0056]\\ =2.25\end{array}$

Therefore, $\text{pH}$ of the solution is $2.25$.

(c)

Interpretation Introduction

Interpretation:

Balanced chemical equation for the reaction of phosphoric acid solution with calcium nitrate resulting in formation of a white precipitate has to be written and the mass of precipitate has to be calculated.

Explanation of Solution

Phosphoric acid solution reacts with aqueous calcium nitrate to form calcium phosphate.  The formed calcium phosphate is precipitated as white precipitate.  Chemical equation can be given as,

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

Balancing calcium atom: In the reactant side, there is only one calcium atom while in the product side, there are three calcium atoms.  Hence, coefficient “3” has to be added before calcium nitrate in the reactant side.  The chemical equation obtained is,

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

Balancing phosphorus atom: In the reactant side, there is only one phosphorus atom while in the product side, there are two phosphorus atoms.  Hence, coefficient “2” has to be added before phosphoric acid in the reactant side.  The chemical equation obtained is,

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

Balancing nitrogen atom: In the reactant side, there are six nitrogen atoms while in the product side, there is one nitrogen atom.  Hence, coefficient “6” has to be added before nitric acid in the product side.  This balances out all the atoms present in the chemical equation.  The balanced chemical equation obtained is,

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

Mass of the precipitate obtained can be calculated by finding the limiting reactant.  In the problem statement it is given that that limiting reactant is phosphoric acid.  Therefore, the mass of precipitate can be calculated as,

  $\begin{array}{c}\text{0}\text{.1614}\text{mol}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{×}\frac{\text{1}\text{mol}{\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}}{\text{2}\text{mol}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}\text{×}\frac{\text{310}\text{.1768}\text{g}{\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}}{\text{1}\text{mol}{\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}}=\frac{50.062}{2}{\text{g Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}}_{\text{)}}\\ \text{=25}{\text{.031g Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}}_{\text{)}}\\ \text{=25}\text{g}{\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}}_{\text{)}}\end{array}$

Therefore, the mass of precipitate is $\text{25}\text{.0}\text{g}$.

(d)

Interpretation Introduction

Interpretation:

Gas that is evolved on addition of zinc to the remaining solution has to be identified and the volume of gas has to be calculated at STP.

Explanation of Solution

Phosphoric acid solution reacts with aqueous calcium nitrate to form calcium phosphate.  The formed calcium phosphate is precipitated as white precipitate.  Balanced chemical equation can be given as,

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

When zinc is added to the solution, it reacts with the nitric acid to liberate hydrogen gas and metal salt.  The salt formed is zinc nitrate.  The chemical equation can be represented as,

    ${\text{Zn(s)+2HNO}}_{\text{3}}\text{(aq)}\to {\text{Zn(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{(aq)+H}}_{\text{2}}\text{(g)}$

The gas that is liberated is hydrogen.  Mol of hydrogen that is liberated can be calculated as shown below,

  $\begin{array}{c}\text{0}\text{.1614}\text{mol}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{×}\frac{\text{6}\text{mol}{\text{HNO}}_{\text{3}}}{\text{2}\text{mol}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}\text{×}\frac{\text{1}\text{mol}{\text{H}}_{\text{2}}}{\text{2}\text{mol}{\text{HNO}}_{\text{3}}}=\frac{0.9684}{4}{\text{mol H}}_{\text{2}}\\ \text{=0}\text{.242 mol}{\text{H}}_{\text{2}}\end{array}$

Volume of hydrogen gas liberated at STP can be calculated using the molar volume of gas as shown below,

    $\text{0}\text{.242 mol}{\text{H}}_{\text{2}}\text{×}\frac{\text{22}\text{.414}\text{L}}{\text{1 mol}{\text{H}}_{\text{2}}}=5.42{\text{L H}}_{\text{2}}$

Therefore, the volume of hydrogen gas liberated is $\text{5}\text{.42}\text{L}$.