Chapter 8, Problem 8.122QA

Interpretation Introduction

To find:

a) Net ionic equation for the reaction between hydroxyapatite ${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2$ and sodium fluoride that produces fluorapatite,${Ca}_{10}{\left(P{O}_4\right)}_6{F}_2$.

b) The concentration of $F^{-}$ in molarity from the given concentration in drinking water i.e.$4 mg/L$.

c) How much fluoride (in mg) is present in a $100 mg$ sample of bone with the given fluoride concentration.

Answer to Problem 8.122QA

Solution:

a) ${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2\left(s\right)+2F^{-}\left(aq\right)\to {Ca}_{10}{\left(P{O}_4\right)}_6{F}_2\left(s\right)+{OH}^{-}\left(aq\right)$

b) The molarity $F^{-}$ is $2.1 \times {10}^{-4} M$

c) $0.600 mg fluoride$

Explanation of Solution

1) Concept:

The net ionic equation is derived from the total ionic equation. The total ionic equation is derived from the complete molecular reaction.

The given concentration of fluoride in $\frac{mg}{L}$ can be converted to moalrity using the molar mass of F and calculating its moles.

2) Formula:

$Molarity=\frac{Moles of solute}{Volume of solution in L}$

3) Given:

Concentration of $F^{-}=4 mg/L$

Fluoride content in bone raised to $6 mg/g$ in 20 years.

Mass of sample of bone = $100 mg.$

4) Calculation:

a)

Net ionic equation for the reaction between hydroxyapatite and sodium fluoride that produces fluorapatite,${Ca}_{10}{\left(P{O}_4\right)}_6{F}_2$.

The reaction equation is,

${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2\left(s\right)+NaF\left(aq\right)\to {Ca}_{10}{\left(P{O}_4\right)}_6{F}_2\left(s\right)+ NaOH\left(aq\right)$

Taking an inventory of atoms on both sides,

$10 Ca+6P+30 O+2H+1Na+1F\to 10 Ca+6P+29 O+1H+1Na+2F$

To balance F-atoms add coefficient in front of $NaF$

${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2\left(s\right)+2NaF\left(aq\right)\to {Ca}_{10}{\left(P{O}_4\right)}_6{F}_2\left(s\right)+ NaOH\left(aq\right)$

Taking an inventory of atoms on both sides,

$10 Ca+6P+30 O+2H+2Na+2F\to 10 Ca+6P+29 O+1H+1Na+2F$

To balance Na-atoms add coefficient in front of $NaOH$

${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2\left(s\right)+2NaF\left(aq\right)\to {Ca}_{10}{\left(P{O}_4\right)}_6{F}_2\left(s\right)+ 2 NaOH\left(aq\right)$

Taking an inventory of atoms on both sides,

$10 Ca+6P+30 O+2H+2Na+2F\to 10 Ca+6P+30 O+2H+2Na+2F$

Therefore, the balanced molecular equation is

${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2\left(s\right)+2NaF\left(aq\right)\to {Ca}_{10}{\left(P{O}_4\right)}_6{F}_2\left(s\right)+ 2 NaOH\left(aq\right)$

Total ionic equation is,

${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2\left(s\right)+2{Na}^{+}\left(aq\right)+2F^{-}\left(aq\right)\to {Ca}_{10}{\left(P{O}_4\right)}_6{F}_2\left(s\right)+ 2{Na}^{+}\left(aq\right)+{OH}^{-}\left(aq\right)$

Cancel out spectator ions,

${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2\left(s\right)+2{Na}^{+}\left(aq\right)+2F^{-}\left(aq\right)\to {Ca}_{10}{\left(P{O}_4\right)}_6{F}_2\left(s\right)+ 2{Na}^{+}\left(aq\right)+{OH}^{-}\left(aq\right)$

Net ionic equation is,

${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2\left(s\right)+2F^{-}\left(aq\right)\to {Ca}_{10}{\left(P{O}_4\right)}_6{F}_2\left(s\right)+{OH}^{-}\left(aq\right)$

b)

Express the concentration of $F^{-}$ in molarity from the given concentration in drinking water i.e.$4 mg/L$.

First convert the concentration mg/L to g/L

$\frac{4 mg}{ L} \times \frac{1 g}{{10}^{3}mg }=4.0 \times {10}^{-3}\frac{g}{L}{F}^{-}$

Convert grams to moles using molar mass

$\frac{4.0 \times {10}^{-3}g}{L} \times \frac{1 mol}{18.998 g}=2.1 \times {10}^{-4}\frac{mol}{L}{F}^{-}$

So the molarity $F^{-}$ is $2.1 \times {10}^{-4} M$

c)

Fluoride (in mg) present in a 100 mg sample of bone:

Fluoride content in bone is  $6\frac{mg}{g}$ i.e. 1 g sample of bone contains 6 mg of Fluoride.

$100 mg bone \times \frac{1 g bone}{{10}^{3}mg bone} \times \frac{6 mg fluoride}{1 g bone}=0.60 mg fluoride$

Conclusion:

The net ionic equation is determined by cancelling the spectator ions in the total ionic equation. Fluoride concentration in the required units is determined using molarity and molar mass of fluoride.