Chapter 17, Problem 132CP

(a)

Interpretation Introduction

Interpretation:

The amount of NaCl, KCl, CaCl₂. 2H₂O, and NaC₃H₅O₃ needed to prepare 100 mL lactated ringer’s solution needs to be determined.

Concept Introduction:

Ringer’s solution is an isotonic solution to blood and it is used in intervenors injection.

The solution which have same osmolarity, same solute concentration, as another solution is known as isotonic solution.

Explanation of Solution

Given information:

285 -315 mg Na+

14.1 -17.3 mg K+

4.9 − 6.0 mg Ca²+

368 − 408 mg Cl-

231 − 261 mg lactate, C₃H₅O₃-

The range is given, first take the average of two values to calculate the approximate amount.

  $m_{N{a}^{+}}=\frac{315+285}{2}=\frac{600}{2}=300\text{ mg}$

Or,

  $m_{K^{+}}=\frac{14.1+17.3}{2}=\frac{31.4}{2}=15.7\text{ mg}$

Or,

  $m_{C{a}^{+}}=\frac{4.9+6.0}{2}=\frac{10.9}{2}=5.45\text{ mg}$

Also,

  $m_{C{l}^{-}}=\frac{368+408}{2}=\frac{776}{2}=388\text{ mg}$

And,

  $m_{{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_3^{-}}=\frac{231+261}{2}=\frac{492}{2}=246\text{ mg}$

Now, the amount of given compounds formed from calculated mass of ions can be calculated as follows:

  ${\text{CaCl}}_{\text{2}}{\text{.2H}}_{\text{2}}\text{O}$ contains 1 mol of Ca atom thus, from 5.45 mg of $C{a}^{2+}$ , the amount of ${\text{CaCl}}_{\text{2}}{\text{.2H}}_{\text{2}}\text{O}$ can be calculated as follows:

Number of moles of $C{a}^{2+}$ atom in 5.45 mg will be:

  $n=\frac{m}{M}=\frac{5.45\text{ mg}}{40.08\text{ g/mol}}=0.136\text{ mol}$

Thus, the number of moles of ${\text{CaCl}}_{\text{2}}{\text{.2H}}_{\text{2}}\text{O}$ formed will be 0.136 mmol. Molar mass of ${\text{CaCl}}_{\text{2}}{\text{.2H}}_{\text{2}}\text{O}$ is 147 g/mol thus, mass of ${\text{CaCl}}_{\text{2}}{\text{.2H}}_{\text{2}}\text{O}$ will be:

  $m=n\times M=\left(0.136\text{ mmol}\right)\left(147\text{ g/mol}\right)=20\text{ mg}$

Now, the mass of $K^{+}$ is 15.7 mg and molar mass is 39.10 g/mol thus, number of moles will be:

  $n=\frac{m}{M}=\frac{15.7\text{ mg}}{39.10\text{ g/mol}}=0.4\text{ mmol}$

The number of moles of KCl formed is 0.4 mmol.

The molar mass of KCl is 74.55 KCl thus, mass of KCl will be:

  $m=n\times M=\left(0.4\text{ mmol}\right)\left(74.55\text{ g/mol}\right)=29.9\text{ mg}$

Similarly, the mass of ${\text{NaC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}$ from 246 mg of ${\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{-}$ can be calculated as follows:

  $m=\left(\frac{2.46\text{ mg}}{89.07\text{ g/mol}}\right)\left(112.06\text{ g/mol}\right)=309.5\text{ mg}$

Here, the mass of sodium ion will be:

  $m_{N{a}^{+}}=m_{{\text{NaC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}}-m_{{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{-}}=\left(309.5-246\right)\text{ mg}=63.5\text{ mg}$

The mass of ${\text{Na}}^{+}$ required from 300 mg is 246 mg.

The mass of NaCl formed from ${\text{Na}}^{+}$ can be calculated as follows:

  $m_{NaCl}=\frac{246\text{ mg }}{22.99\text{ g/mol}}\left(58.44\text{ g/mol}\right)=601.2\text{ mg}$

The amount of ${\text{Cl}}^{-}$ from ${\text{CaCl}}_{\text{2}}.2{\text{H}}_{\text{2}}\text{O}$ can be calculated as follows:

  $m=\frac{20\text{ mg}}{147\text{ g/mol}}\times 2\times 35.5\text{ g/mol=9}\text{.6 mg}$

It can also be calculated from mass of KCl and NaCl as follows:

From KCl:

  $m_{{\text{ Cl}}^{-}}=29.9\left(KCl\right)-15.7\left(K^{+}\right)=14.2\text{ mg}$

Similarly,

  $m_{{\text{ Cl}}^{-}}=601.2\left(NaCl\right)-236.5\left(N{a}^{+}\right)=364.7\text{ mg}$

Now, the total mass of chloride ion will be:

  $m_{C{l}^{-}}=9.6\text{ mg+14}\text{.2 mg+364}\text{.7 mg=388}\text{.5 mg}$

(b)

Interpretation Introduction

Interpretation:

The range of the osmotic pressure of the solution at 37⁰C should be predicted.

Concept Introduction:

Ringer’s solution is an isotonic solution to blood and it is used in intervenors injection.

The solution which have same osmolarity, same solute concentration, as another solution is known as isotonic solution.

Answer to Problem 132CP

Range of osmotic pressure = (4.38 − 6.55)

Explanation of Solution

The solution contains NaCl + KCl+ CaCl₂ + NaC₃H₅O₃

The dissociation reaction is shown below:

  $\begin{array}{c}\text{NaCl}\rightleftarrows {\text{Na}}^{\text{+}}\text{+}{\text{Cl}}^{\text{-}}\\ \text{KCl}\rightleftarrows {\text{K}}^{\text{+}}\text{+}{\text{Cl}}^{\text{-}}\\ {\text{CaCl}}_{\text{2}}\rightleftarrows {\text{Ca}}^{\text{2+}}\text{+}{\text{2Cl}}^{\text{-}}\end{array}$

  $Na{C}_3{H}_5{O}_3\rightleftharpoons N{a}^{+}+C_3{H}_5{O}_3{}^{-}$

The total molarity will be sum of molarity of all the solutions.

For the initial range of the osmotic pressure, the total molarity should be calculated by taking the initial value of amount of ions from their given range.

The calculation of molarity is shown below:

  $\begin{array}{c}{\text{M}}_{{\text{Na}}^{\text{+}}}\text{=}\frac{\text{285}\text{×}{\text{10}}^{\text{-3}}}{\text{23}\text{×}\text{100}\text{×}{\text{10}}^{\text{-3}}}\\ \text{=}\text{0}\text{.123}\text{M}\\ {\text{M}}_{{\text{K}}^{\text{+}}}\text{=}\frac{\text{14}\text{.1}\text{×}{\text{10}}^{\text{-3}}}{\text{39}\text{×}\text{100}\text{×}{\text{10}}^{\text{-3}}}\\ \text{=}\text{3}\text{.615}\text{×}{\text{10}}^{\text{-3}}\text{M}\\ {\text{M}}_{{\text{Ca}}^{\text{2+}}}\text{=}\frac{\text{4}\text{.9}\text{×}{\text{10}}^{\text{-3}}}{\text{40}\text{×}\text{100}\text{×}{\text{10}}^{\text{-3}}}\\ \text{=}\text{1}\text{.225}\text{×}{\text{10}}^{\text{-3}}\text{M}\\ {\text{M}}_{\left({\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{-}\right)}\text{=}\frac{\text{231}\text{×}{\text{10}}^{\text{-3}}}{\text{89}\text{×}\text{100}\text{×}{\text{10}}^{\text{-3}}}\\ \text{=}\text{0}\text{.0259}\text{M}\\ {\text{M}}_{{\text{Cl}}^{\text{-}}}\text{=}\frac{\text{368}\text{×}{\text{10}}^{\text{-3}}}{\text{35}\text{.5}\text{×}\text{100}\text{×}{\text{10}}^{\text{-3}}}\\ \text{=}\text{0}\text{.1036}\text{M}\end{array}$

The calculation of osmotic pressure is shown below:

  $\begin{array}{c}{\text{π}}_{\text{1}}\text{=}\text{MRT}\\ \text{=}\text{RT}\left({\text{M}}_{{\text{Na}}^{\text{+}}}\text{+}{\text{M}}_{{\text{K}}^{\text{+}}}\text{+}{\text{M}}_{\left({\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{-}\right)}\text{+}{\text{M}}_{{\text{Cl}}^{\text{-}}}\text{+}{\text{M}}_{{\text{Ca}}^{\text{2+}}}\right)\\ \text{=}\text{0}\text{.0821}\text{×}\text{310}\left(\text{0}\text{.123}\text{+}\text{3}\text{.615}{\text{×10}}^{\text{-3}}\text{+}\text{1}\text{.225}{\text{×10}}^{\text{-3}}\text{+}\text{0}\text{.0259}\text{+0}\text{.1036}\right)\\ \text{=}\text{6}\text{.55}\text{atm}\end{array}$

Therefore, the minimum range of the osmotic pressure is 6.55 atm.

Now, for maximum range, take the final value of mass from the given ranges of masses of ions.

The calculation of molarity is shown below:

  $\begin{array}{c}{\text{M}}_{{\text{Na}}^{\text{+}}}\text{=}\frac{\text{315}\text{×}{\text{10}}^{\text{-3}}}{\text{23}\text{×}\text{100}\text{×}{\text{10}}^{\text{-3}}}\\ \text{=}\text{0}\text{.1369}\text{M}\\ {\text{M}}_{{\text{K}}^{\text{+}}}\text{=}\frac{\text{17}\text{.3}\text{×}{\text{10}}^{\text{-3}}}{\text{39}\text{×}\text{100}\text{×}{\text{10}}^{\text{-3}}}\\ \text{=}\text{4}\text{.435}\text{×}{\text{10}}^{\text{-3}}\text{M}\\ {\text{M}}_{{\text{Ca}}^{\text{2+}}}\text{=}\frac{\text{6}\text{×}{\text{10}}^{\text{-3}}}{\text{40}\text{×}\text{100}\text{×}{\text{10}}^{\text{-3}}}\\ \text{=}\text{1}\text{.5}\text{×}{\text{10}}^{\text{-3}}\text{M}\\ {\text{M}}_{\left({\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{-}\right)}\text{=}\frac{\text{261}\text{×}{\text{10}}^{\text{-3}}}{\text{89}\text{×}\text{100}\text{×}{\text{10}}^{\text{-3}}}\\ \text{=}\text{0}\text{.0293}\text{M}\\ {\text{M}}_{{\text{Cl}}^{\text{-}}}\text{=}\frac{\text{408}\text{×}{\text{10}}^{\text{-3}}}{\text{35}\text{.5}\text{×}\text{100}\text{×}{\text{10}}^{\text{-3}}}\\ \text{=}\text{0}\text{.1149}\text{M}\end{array}$

The calculation of osmotic pressure is shown below:

  $\begin{array}{c}{\text{π}}_{\text{2}}\text{=}\text{MRT}\\ \text{=}\text{RT}\left({\text{M}}_{{\text{Na}}^{\text{+}}}\text{+}{\text{M}}_{{\text{K}}^{\text{+}}}\text{+}{\text{M}}_{\left({\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}\right)}\text{+}{\text{M}}_{{\text{Cl}}^{\text{-}}}\text{+}{\text{M}}_{{\text{Ca}}^{\text{2+}}}\right)\\ =0.0821\times 310(0.1369+4.435\times {10}^{-3}+1.5\times {10}^{-3}+0.0293+0.1149)\\ =7.30atm\end{array}$

Range of osmotic pressure = (6.55-7.30 atm)