Chapter 10, Problem 108AE

Interpretation Introduction

Interpretation: The amount of reagents required and range of osmotic pressure has to be calculated.

Concept Introduction: The mass of the compound is calculated by taking the products of molar mass of the compound to the given mass. The mass of compound can be given by,

$\text{Mass}\text{of}\text{compound(in}\text{grams)=Molar}\text{mass}\text{(in}\text{g)×Given}\text{mass(in}\text{g)}$

Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution.

The osmotic pressure can be given by the equation,

$\text{Π=MRT}$

$\begin{array}{l}\text{Here,Π=Osmotic}\text{pressure}\\ \text{M=Molarity}\text{of}\text{solution}\\ \text{R=}\text{Gas}\text{law}\text{constant}\\ \text{T=Temperature}\end{array}$

Answer to Problem 108AE

$\text{305}\text{.9}\text{mg}$ Sodium lactate, $\text{20}\text{.0}\text{mg}{\text{CaCl}}_{\text{2}}{\text{.2H}}_{\text{2}}\text{O}$, $\text{29}\text{.9}\text{mg}\text{KCl}$ and $\text{601}\text{.2}\text{mg}\text{NaCl}$ are required to prepare lactated Ringer’s reagent.

The range of osmotic pressure is $\text{6}\text{.59}\text{atm}\text{to}\text{7}\text{.30}\text{atm}$

Explanation of Solution

Record the given info

Mass of Sodium = $\text{285-315}\text{mg}$

Mass of Potassium = $\text{14}\text{.1-17}\text{.3}\text{mg}$

Mass of Calcium =$\text{4}\text{.9-6}\text{.0}\text{mg}$

Mass of Chlorine =$\text{368-408}\text{mg}$

Mass of Lactate = $\text{231-261}\text{mg}$

To calculate the mass of individual elements

Molar mass of Sodium lactate = $\text{112}\text{.06}\text{mg}$

Molar mass of Lactate = $\text{89}\text{.07}\text{mg}$

Molar mass of ${\text{CaCl}}_{\text{2}}{\text{.2H}}_{\text{2}}\text{O}$= $\text{147}\text{.01}\text{mg}$

Molecular mass of Calcium = $\text{40}\text{.08}\text{mg}$

Molar mass of $\text{KCl}$= $\text{74}\text{.55}\text{mg}$

Molecular weight of Potassium = $\text{39}\text{.10}\text{mg}$

Molar mass of $\text{NaCl}$ = $\text{58}\text{.44}\text{mg}$

Molecular mass of Sodium= $\text{22}\text{.99}\text{mg}$

The average values for each ion are,

$\text{300}\text{.00}\text{mg}{\text{Na}}^{\text{+}}\text{,15}\text{.7}\text{mg}{\text{K}}^{\text{+}}\text{,5}\text{.45}\text{mg}{\text{Ca}}^{\text{2+}}\text{,388}\text{mg}{\text{Cl}}^{\text{-}}\text{and}\text{246}\text{mg}\text{Lactate}$

The source of Lactate is ${\text{NaC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}$

Mass of Lactate = $\text{246}\text{mg}{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{\text{-}}\text{×}\frac{\text{112}\text{.06}\text{mg}{\text{NaC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}}{\text{89}\text{.07}\text{mg}{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{\text{-}}}\text{=309}\text{.5}\text{mg}$

The source of ${\text{Ca}}^{\text{2+}}$ is ${\text{CaCl}}_{\text{2}}{\text{.2H}}_{\text{2}}\text{O}$

 Mass of ${\text{CaCl}}_{\text{2}}{\text{.2H}}_{\text{2}}\text{O}$= $\text{5}\text{.45}\text{mg}{\text{Ca}}^{\text{2+}}\text{×}\frac{\text{147}\text{.01}\text{mg}{\text{CaCl}}_{\text{2}}{\text{.2H}}_{\text{2}}\text{O}}{\text{40}\text{.08}\text{mg}{\text{Ca}}^{\text{2+}}}\text{=20}\text{mg}{\text{CaCl}}_{\text{2}}{\text{.2H}}_{\text{2}}\text{O}$

The source of ${\text{K}}^{\text{+}}$ is $\text{KCl}$

Mass of $\text{KCl}$= $\text{15}\text{.7}\text{mg}{\text{K}}^{\text{+}}\text{×}\frac{\text{74}\text{.55}\text{mg}\text{KCl}}{\text{39}\text{.10}\text{mg}{\text{K}}^{\text{+}}}\text{=29}\text{.9}\text{mg}$

Mass of ${\text{Na}}^{\text{+}}$ added =$\text{309}\text{.5}\text{mg}\text{Sodium}\text{lactate-246}\text{.0}\text{mg}\text{Lactate=63}\text{.5}\text{mg}{\text{Na}}^{\text{+}}$

Additional amount of Sodium $\text{236}\text{.5}\text{mg}{\text{Na}}^{\text{+}}$ is added to get desired $\text{300}\text{mg}$

Mass of Sodium added =$\text{236}\text{.5}\text{mg}{\text{Na}}^{\text{+}}\text{×}\frac{\text{58}\text{.44}\text{mg}\text{NaCl}}{\text{22}\text{.99}\text{mg}{\text{Na}}^{\text{+}}}\text{=601}\text{.2}\text{mg}$

Mass of ${\text{Cl}}^{\text{-}}$ added = $\text{20}\text{.0}\text{mg}{\text{CaCl}}_{\text{2}}{\text{.2H}}_{\text{2}}\text{O×}\frac{\text{70}\text{.90}\text{mg}{\text{Cl}}^{\text{-}}}{\text{147}\text{.01}\text{mg}{\text{CaCl}}_{\text{2}}{\text{.2H}}_{\text{2}}\text{O}}\text{=9}\text{.6}\text{mg}$

$\begin{array}{l}\text{20}\text{.0}\text{mg}{\text{CaCl}}_{\text{2}}{\text{.2H}}_{\text{2}}\text{O}\text{=9}\text{.6}\text{mg}{\text{Cl}}^{\text{-}}\\ \text{29}\text{.9}\text{mg}\text{KCl-15}\text{.7}\text{mg}{\text{K}}^{\text{+}}\text{=14}\text{.2}\text{mg}{\text{Cl}}^{\text{-}}\\ \text{601}\text{.2}\text{mg}\text{NaCl-236}\text{.5}\text{mg}{\text{Na}}^{\text{+}}\text{=364}\text{.7}\text{mg}{\text{Cl}}^{\text{-}}\end{array}$

                    Total ${\text{Cl}}^{\text{-}}$          =$\text{388}\text{.5}\text{mg}$

Therefore,

$\text{305}\text{.9}\text{mg}$ Sodium lactate, $\text{20}\text{.0}\text{mg}{\text{CaCl}}_{\text{2}}{\text{.2H}}_{\text{2}}\text{O}$, $\text{29}\text{.9}\text{mg}\text{KCl}$ and $\text{601}\text{.2}\text{mg}\text{NaCl}$ are required to prepare lactated Ringer’s reagent.

Record the given info

Mass of Sodium = $\text{285-315}\text{mg}$

Mass of Potassium = $\text{14}\text{.1-17}\text{.3}\text{mg}$

Mass of Calcium =$\text{4}\text{.9-6}\text{.0}\text{mg}$

Mass of Chlorine =$\text{368-408}\text{mg}$

Mass of Lactate = $\text{231-261}\text{mg}$

To calculate the minimum and maximum concentrations of ions

Molar mass of Lactate = $\text{89}\text{.07}\text{mg}$

Molecular mass of Calcium = $\text{40}\text{.08}\text{mg}$

Molecular weight of Potassium = $\text{39}\text{.10}\text{mg}$

Molecular mass of Sodium= $\text{22}\text{.99}\text{mg}$

At minimum concentration,

Molarity of Sodium = $\frac{\text{285}\text{mg}{\text{Na}}^{\text{+}}}{\text{100}\text{.}\text{mL}}\text{×}\frac{\text{1}\text{mmol}}{\text{22}\text{.99}\text{mg}}\text{=0}\text{.124}\text{M}$

Molarity of Potassium = $\frac{\text{14}\text{.1}\text{mg}{\text{K}}^{\text{+}}}{\text{100}\text{.mL}}\text{×}\frac{\text{1}\text{mmol}}{\text{39}\text{.10}\text{mg}}\text{=0}\text{.00361}\text{M}$

Molarity of Lactate = $\frac{\text{231}\text{mg}{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{\text{-}}}{\text{100}\text{.mL}}\text{×}\frac{\text{1}\text{mmol}}{\text{89}\text{.07}\text{mg}}\text{=0}\text{.0259}\text{M}$

Molarity of Calcium = $\frac{\text{4}\text{.9}\text{mg}{\text{Ca}}^{\text{2+}}}{\text{100}\text{.mL}}\text{×}\frac{\text{1}\text{mmol}}{\text{40}\text{.08}\text{mg}}\text{=0}\text{.0012}\text{M}$

Molarity of Chlorine = $\frac{\text{368}\text{mg}{\text{Cl}}^{\text{-}}}{\text{100}\text{.mL}}\text{×}\frac{\text{1}\text{mmol}}{\text{35}\text{.45}\text{mg}}\text{=0}\text{.104}\text{M}$

The total concentration = $\text{0}\text{.124+0}\text{.00361+0}\text{.0012+0}\text{.104+0}\text{.0259}$

                                       = $\text{0}\text{.259}\text{M}$

At maximum concentration,

Molarity of Sodium = $\frac{\text{315}\text{mg}{\text{Na}}^{\text{+}}}{\text{100}\text{.}\text{mL}}\text{×}\frac{\text{1}\text{mmol}}{\text{22}\text{.99}\text{mg}}\text{=0}\text{.137}\text{M}$

Molarity of Potassium = $\frac{\text{17}\text{.3}\text{mg}{\text{K}}^{\text{+}}}{\text{100}\text{.mL}}\text{×}\frac{\text{1}\text{mmol}}{\text{39}\text{.10}\text{mg}}\text{=0}\text{.00442}\text{M}$

Molarity of Lactate = $\frac{\text{261}\text{mg}{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{\text{-}}}{\text{100}\text{.mL}}\text{×}\frac{\text{1}\text{mmol}}{\text{89}\text{.07}\text{mg}}\text{=0}\text{.0293}\text{M}$

Molarity of Calcium = $\frac{6.0\text{mg}{\text{Ca}}^{\text{2+}}}{\text{100}\text{.mL}}\text{×}\frac{\text{1}\text{mmol}}{\text{40}\text{.08}\text{mg}}\text{=0}\text{.0015}\text{M}$

Molarity of Chlorine = $\frac{\text{408}\text{mg}{\text{Cl}}^{\text{-}}}{\text{100}\text{.mL}}\text{×}\frac{\text{1}\text{mmol}}{\text{35}\text{.45}\text{mg}}\text{=0}\text{.115}\text{M}$

The total concentration= $\text{0}\text{.137+0}\text{.00442+0}\text{.0015+0}\text{.115+0}\text{.0293}$

                                     = $\text{0}\text{.287}\text{M}$

The total concentration of ions for minimum and maximum concentration is calculated by using the summing the molarities of individual ions. The molarities of individual ions are calculated using the minimum and maximum moles to their molecular masses. The total concentrations at minimum and maximum concentrations are $\text{0}\text{.259}\text{M}$ and $\text{0}\text{.287}\text{M}$ respectively.

To calculate the osmotic pressure at minimum and maximum concentration

At minimum concentration,

$\begin{array}{l}\text{π=MRT}\\ \\ \text{=}\frac{\text{0}\text{.259}\text{mol}}{\text{L}}\text{×}\frac{\text{0}\text{.08206}\text{L}\text{atm}}{\text{K}\text{mol}}\text{×310}\text{.K}\\ \\ \text{=6}\text{.59}\text{atm}\end{array}$

At maximum concentration,

$\begin{array}{l}\text{π=MRT}\\ \\ \text{=}\frac{\text{0}\text{.287}\text{mol}}{\text{L}}\text{×}\frac{\text{0}\text{.08206}\text{L}\text{atm}}{\text{K}\text{mol}}\text{×310}\text{.K}\\ \\ \text{=7}\text{.30}\text{atm}\\ \end{array}$

At minimum concentration, osmotic pressure= $\text{6}\text{.59}\text{atm}$

At maximum concentration, osmotic pressure= $\text{7}\text{.30}\text{atm}$

Conclusion

The mass of individual elements was calculated using their respective molar mass and molecular weight and the given weight. A typical analytical balance can nearly measure to $\text{0}\text{.1}\text{mg}$. Therefore, $\text{305}\text{.9}\text{mg}$ Sodium lactate, $\text{20}\text{.0}\text{mg}{\text{CaCl}}_{\text{2}}{\text{.2H}}_{\text{2}}\text{O}$, $\text{29}\text{.9}\text{mg}\text{KCl}$ and $\text{601}\text{.2}\text{mg}\text{NaCl}$ are required to prepare lactated Ringer’s reagent.

The osmotic pressure at minimum and maximum concentrations was calculated using the molarities at minimum and maximum concentration. The osmotic pressure at minimum and maximum concentrations were $\text{6}\text{.59}\text{atm}$ and $\text{7}\text{.30}\text{atm}$.