Chapter F, Problem I.26E

(a)

Interpretation Introduction

Interpretation:

Chemical equation, complete ionic equation and also the net ionic equation for the reaction between ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}$ and ${\text{CaCl}}_{\text{2}}$ has to be written.

Explanation of Solution

The reaction given is between aqueous solutions of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}$ and ${\text{CaCl}}_{\text{2}}$.  The reaction can be given as follows:

    ${\text{2(NH}}_{\text{4}}{\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}(\text{aq})+3{\text{CaCl}}_{\text{2}}(\text{aq})\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+{\text{6NH}}_{\text{4}}\text{Cl(aq)}$

Aqueous solution of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}$ reacts with ${\text{CaCl}}_{\text{2}}$ resulting in formation of ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ and ${\text{NH}}_{\text{4}}\text{Cl}$.  The precipitate that is formed is ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$.

The balanced chemical equation for the reaction is given as follows:

    ${\text{2(NH}}_{\text{4}}{\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}(\text{aq})+3{\text{CaCl}}_{\text{2}}(\text{aq})\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+{\text{6NH}}_{\text{4}}\text{Cl(aq)}$

Complete ionic equation is written by showing the ions present in aqueous solution.  This is given as follows:

${\text{6NH}}_{\text{4}}{}^{\text{+}}{\text{(aq)+2PO}}_{\text{4}}{}^{3-}(\text{aq})+3{\text{Ca}}^{\text{2+}}{\text{(aq)+6Cl}}^{-}(\text{aq})\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+{\text{6NH}}_{\text{4}}{}^{\text{+}}{\text{(aq)+6Cl}}^{-}\text{(aq)}$

In the complete ionic equation, the common ions that are present on both sides of the equation is known as spectator ions.  Thus in this case, the spectator ions are ${\text{NH}}_{\text{4}}{}^{\text{+}}$ and ${\text{Cl}}^{-}$.  Net ionic equation can be obtained from the complete ionic equation by removing the spectator ions.

$\cancel{{\text{6NH}}_{\text{4}}{}^{\text{+}}\text{(aq)}}{\text{+2PO}}_{\text{4}}{}^{3-}(\text{aq})+3{\text{Ca}}^{\text{2+}}\text{(aq)+}\cancel{{\text{6Cl}}^{-}(\text{aq})}\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+\cancel{{\text{6NH}}_{\text{4}}{}^{\text{+}}\text{(aq)}}\text{+}\cancel{{\text{6Cl}}^{-}\text{(aq)}}$

Thus net ionic equation can be written as follows:

    ${\text{2PO}}_{\text{4}}{}^{3-}(\text{aq})+3{\text{Ca}}^{\text{2+}}\text{(aq)}\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)$

(b)

Interpretation Introduction

Interpretation:

Concentration of spectator ions in the final solution with volume of $\text{70}\text{.0}\text{mL}$ obtained when $\text{2}\text{.50}\text{g}$ of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}$ is added to $\text{50}\text{.0}\text{mL}$ of $\text{0}\text{.125}\text{M}{\text{CaCl}}_{\text{2}}$.

Answer to Problem I.26E

Concentration of ammonium ions is $0.72\text{mol}\cdot {\text{L}}^{-1}$ and chloride ions is $0.179\text{mol}\cdot {\text{L}}^{-1}$.

Explanation of Solution

Concentration of ammonium ions:

Number of moles of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}$ is calculated as shown below:

    $\begin{array}{c}{n}_{{{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}}=\frac{m}{M}\\ =\frac{2.50\text{g}}{149.09\text{g}\cdot {\text{mol}}^{-1}}\\ =0.0168\text{mol}\end{array}$

One mole of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}$ contains three moles of ${\text{NH}}_{\text{4}}{}^{\text{+}}$ ions.  Therefore, $\text{0}\text{.0168}\text{mol}$ of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}$ contains:

    $\begin{array}{c}{n}_{{\text{NH}}_{\text{4}}{}^{\text{+}}}=\text{0}\text{.0168}\text{mol}{{\text{(NH}}_{\text{4}}}_{\text{)}}{\text{PO}}_{\text{4}}\times \frac{\text{6}\text{mol}{\text{NH}}_{\text{4}}{}^{\text{+}}\text{ion}}{\text{2}\text{mol}{{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}}\\ =0.0504\text{mol}{\text{NH}}_{\text{4}}{}^{+}\end{array}$

Therefore, the concentration of ammonium ions in $\text{70}\text{.0}\text{mL}$ of final solution can be calculated as follows:

    $\begin{array}{c}M=\frac{n}{V}\\ =\frac{\text{0}\text{.0504}\text{mol}}{\text{0}\text{.07}\text{L}}\\ =0.72\text{mol}\cdot {\text{L}}^{-1}\end{array}$

Thus, the concentration of ammonium ions is $0.72\text{mol}\cdot {\text{L}}^{-1}$.

Concentration of chloride ions:

The concentration of the solution is $\text{0}\text{.125}\text{M}$ of ${\text{CaCl}}_{\text{2}}$ and volume is $\text{50}\text{.0}\text{mL}$.

One mole of ${\text{CaCl}}_{\text{2}}$ contains two moles of ${\text{Cl}}^{-}$ ions.  Therefore, $\text{0}\text{.125}\text{M}$ of ${\text{CaCl}}_{\text{2}}$ contains $\text{0}\text{.250}\text{M}$ of ${\text{Cl}}^{-}$ ions.  Given volume is $\text{50}\text{.0}\text{mL}$.  Therefore, the moles of chloride ions in $\text{50}\text{.0}\text{mL}$ is calculated as follows:

    $\begin{array}{c}n=0.25\text{mol}\cdot {\text{L}}^{-1}\times 0.0500\text{L}\\ =0.0125\text{mol}\end{array}$

The number of moles of ${\text{Cl}}^{-}$ ion is $\text{0}\text{.0125}\text{mol}$ and volume is $\text{0}\text{.070}\text{L}$.  Therefore, the concentration of chloride ions can be calculated as follows:

    $\begin{array}{c}M=\frac{n}{V}\\ =\frac{\text{0}\text{.0125}\text{mol}}{\text{0}\text{.07}\text{L}}\\ =0.179\text{mol}\cdot {\text{L}}^{-1}\end{array}$

Thus, the concentration of chloride ions is $0.179\text{mol}\cdot {\text{L}}^{-1}.$