Colligative Properties in Pharmaceutical Solutions

1

Vapor Pressure of Glucose Solution

Calculate the vapor pressure above a glucose solution where the mole fraction of solute is 0.0098. The vapor pressure of the pure solvent is 2.995 kPa.

Step-by-Step Solution:

Identify given values:

Mole fraction of solute (\(X_{\text{solute}}\)) = 0.0098
Vapor pressure of pure solvent (\(P^\circ_{\text{solvent}}\)) = 2.995 kPa

Calculate mole fraction of solvent:

\(X_{\text{solvent}} = 1 - X_{\text{solute}} = 1 - 0.0098 = 0.9902\)

Since glucose is a non-volatile solute, it reduces the mole fraction of solvent molecules at the surface.

Apply Raoult's Law:

\(P_{\text{solution}} = X_{\text{solvent}} \times P^\circ_{\text{solvent}}\)

\(P_{\text{solution}} = 0.9902 \times 2.995 = 2.966 \text{kPa}\)

The vapor pressure of the glucose solution is 2.966 kPa

2

Osmotic Pressure for KCl Solution

How many grams of KCl must be dissolved in 100 mL of water at 37°C to prepare an isotonic solution with an osmotic pressure of 5.6 atm?

Step-by-Step Solution:

Convert temperature to Kelvin:

\(T = 37 + 273 = 310 \text{K}\)

Determine van't Hoff factor for KCl:

KCl dissociates completely in water: \(\text{KCl} \rightarrow \text{K}^+ + \text{Cl}^-\)

\(i = 2\)

The van't Hoff factor accounts for the number of particles produced per formula unit dissolved.

Use osmotic pressure formula:

\(\pi = iCRT\)

Rearrange to solve for molarity (C):

\(C = \frac{\pi}{iRT} = \frac{5.6}{2 \times 0.0821 \times 310}\)

\(C = 0.110 \text{mol/L}\)

Calculate moles of KCl in 100 mL solution:

\(n = C \times V = 0.110 \frac{\text{mol}}{\text{L}} \times 0.100 \text{L} = 0.0110 \text{mol}\)

Convert moles to grams:

Molar mass of KCl = 39.10 + 35.45 = 74.55 g/mol

\(\text{Mass} = n \times \text{molar mass} = 0.0110 \times 74.55 = 0.820 \text{g}\)

0.820 grams of KCl is required

3

Molality of Potassium Citrate Solution

Calculate the molality of a potassium citrate solution that boils at 100.067°C. Given: \(K_b = 0.512 ^\circ\text{C·kg/mol}\) and van't Hoff factor \(i = 3.8\).

Step-by-Step Solution:

Calculate boiling point elevation:

Normal boiling point of water = 100.000°C

\(\Delta T_b = T_{\text{solution}} - T_{\text{pure}} = 100.067 - 100.000 = 0.067 ^\circ\text{C}\)

Apply boiling point elevation formula:

\(\Delta T_b = i \cdot K_b \cdot m\)

Rearrange to solve for molality (m):

\(m = \frac{\Delta T_b}{i \cdot K_b} = \frac{0.067}{3.8 \times 0.512}\)

Calculate the molality:

\(m = \frac{0.067}{1.9456} = 0.0344 \text{mol/kg}\)

Potassium citrate (\(K_3C_6H_5O_7\)) partially dissociates into 3.8 particles on average in solution, which is why i = 3.8.

The molality is 0.0344 mol/kg

4

Freezing Point of PABA Solution

At what temperature will 200 g of a 0.007% aqueous solution of para-aminobenzoic acid (PABA) freeze? (\(K_f = 1.86 ^\circ\text{C·kg/mol}\))

Step-by-Step Solution:

Interpret the concentration:

0.007% solution means 0.007 g PABA per 100 g solution

Assume 200 g solution contains 0.014 g PABA (0.007% of 200 g)

Solvent mass ≈ 200 g - 0.014 g = 199.986 g ≈ 0.200 kg

Calculate molar mass of PABA (C₇H₇NO₂):

\(M = 7(12.01) + 7(1.01) + 14.01 + 2(16.00) = 137.14 \text{g/mol}\)

Calculate moles of PABA:

\(n = \frac{0.014}{137.14} = 1.021 \times 10^{-4} \text{mol}\)

Calculate molality:

\(m = \frac{1.021 \times 10^{-4}}{0.200} = 5.105 \times 10^{-4} \text{mol/kg}\)

Calculate freezing point depression:

Assume PABA is a non-electrolyte (\(i = 1\))

\(\Delta T_f = i \cdot K_f \cdot m = 1 \times 1.86 \times 5.105 \times 10^{-4}\)

\(\Delta T_f = 0.00095 ^\circ\text{C}\)

Calculate freezing point:

\(T_f = 0 - \Delta T_f = -0.00095 ^\circ\text{C}\)

The extremely small freezing point depression is due to the very low concentration (0.007%) of the solution, which is typical for ophthalmic preparations to avoid irritation.

The solution freezes at -0.00095°C

5

Molar Mass of Hemoglobin

Determine the molar mass of hemoglobin in a solution with 10 g of hemoglobin in 100 mL at 37°C, with an osmotic pressure of 3.847 kPa.

Step-by-Step Solution:

Convert temperature to Kelvin:

\(T = 37 + 273 = 310 \text{K}\)

Convert osmotic pressure to atm:

\(\pi = 3.847 \text{kPa} \times \frac{1 \text{atm}}{101.325 \text{kPa}} = 0.03797 \text{atm}\)

Use osmotic pressure formula:

\(\pi = \frac{n}{V}RT\)

Rearrange to solve for moles (n):

\(n = \frac{\pi V}{RT}\)

Calculate moles of hemoglobin:

\(n = \frac{0.03797 \text{atm} \times 0.100 \text{L}}{0.0821 \frac{\text{L·atm}}{\text{mol·K}} \times 310 \text{K}} = 1.493 \times 10^{-4} \text{mol}\)

Calculate molar mass:

\(\text{Molar mass} = \frac{\text{mass}}{n} = \frac{10 \text{g}}{1.493 \times 10^{-4} \text{mol}} = 67,000 \text{g/mol}\)

This molar mass is typical for hemoglobin, which is a large protein molecule important for oxygen transport in the blood.

The molar mass of hemoglobin is 67,000 g/mol

6

Molar Mass of Ferritin

Determine the molar mass of ferritin in a 300 mL aqueous solution containing 30 g of dissolved substance, with an osmotic pressure of 0.0057 atm at 37°C.

Step-by-Step Solution:

Convert temperature to Kelvin:

\(T = 37 + 273 = 310 \text{K}\)

Use osmotic pressure formula:

\(\pi = \frac{n}{V}RT\)

Rearrange to solve for moles (n):

\(n = \frac{\pi V}{RT}\)

Calculate moles of ferritin:

\(n = \frac{0.0057 \text{atm} \times 0.300 \text{L}}{0.0821 \frac{\text{L·atm}}{\text{mol·K}} \times 310 \text{K}} = 6.719 \times 10^{-5} \text{mol}\)

Calculate molar mass:

\(\text{Molar mass} = \frac{30 \text{g}}{6.719 \times 10^{-5} \text{mol}} = 446,500 \text{g/mol}\)

Ferritin is a very large protein (molecular weight ~450,000 g/mol) that stores iron in cells. This high molecular weight is typical for storage proteins.

The molar mass of ferritin is 446,500 g/mol

7

Vapor Pressure of Potassium Bicarbonate Solution

Calculate the vapor pressure over a solution containing 15 g of potassium bicarbonate in 200 g of solvent. The vapor pressure over the pure solvent is 29.28 mmHg.

Step-by-Step Solution:

Determine van't Hoff factor for K₂CO₃:

K₂CO₃ dissociates into 3 ions: \(K_2CO_3 \rightarrow 2K^+ + CO_3^{2-}\)

\(i = 3\)

Calculate molar mass of K₂CO₃:

\(M = 2(39.10) + 12.01 + 3(16.00) = 138.21 \text{g/mol}\)

Calculate moles of K₂CO₃:

\(n_{\text{solute}} = \frac{15}{138.21} = 0.1085 \text{mol}\)

Calculate total moles of particles:

\(n_{\text{particles}} = i \times n_{\text{solute}} = 3 \times 0.1085 = 0.3255 \text{mol}\)

Calculate moles of solvent (water):

\(n_{\text{solvent}} = \frac{200}{18.02} = 11.11 \text{mol}\)

Calculate mole fraction of solvent:

\(X_{\text{solvent}} = \frac{11.11}{11.11 + 0.3255} = 0.9715\)

Apply Raoult's Law:

\(P_{\text{solution}} = X_{\text{solvent}} \times P^\circ_{\text{solvent}} = 0.9715 \times 29.28 = 28.45 \text{mmHg}\)

The vapor pressure lowering is due to both the presence of solute and its dissociation into multiple particles, which is important for understanding drug solubility.

The vapor pressure is 28.45 mmHg

8

Glucose Mass for Isotonic Solution

How many grams of glucose (C₆H₁₂O₆) must be dissolved in 500 mL of water to match the osmotic pressure of a solution containing 9.2 g of glycerol in 1 liter?

Step-by-Step Solution:

Calculate molarity of glycerol solution:

Molar mass of glycerol (C₃H₈O₃) = 92.09 g/mol

\(M_{\text{glycerol}} = \frac{9.2}{92.09} = 0.100 \text{mol/L}\)

Set osmotic pressures equal:

For isotonic solutions: \(\pi_{\text{glucose}} = \pi_{\text{glycerol}}\)

Since both are non-electrolytes (i = 1), their molarities must be equal:

\(M_{\text{glucose}} = M_{\text{glycerol}} = 0.100 \text{mol/L}\)

Calculate moles of glucose needed:

\(n = M \times V = 0.100 \frac{\text{mol}}{\text{L}} \times 0.500 \text{L} = 0.0500 \text{mol}\)

Calculate mass of glucose:

Molar mass of glucose = 180.16 g/mol

\(\text{Mass} = 0.0500 \text{mol} \times 180.16 \frac{\text{g}}{\text{mol}} = 9.01 \text{g}\)

Creating isotonic solutions is crucial in pharmaceuticals to prevent cell damage when administering IV fluids or eye drops.

9.01 grams of glucose is required

9

Tonicity Comparison: Glucose vs. Sucrose

Is a 5% glucose solution hypotonic, hypertonic, or isotonic relative to a 5% sucrose solution?

Step-by-Step Solution:

Define the concentrations:

5% solution = 5 g solute per 100 g solution ≈ 5 g solute per 100 mL solution

Assume density ≈ 1 g/mL for both solutions

Calculate molarity of glucose solution:

Molar mass of glucose (C₆H₁₂O₆) = 180.16 g/mol

\(M_{\text{glucose}} = \frac{5}{180.16} \times \frac{1000}{100} = 0.277 \text{M}\)

Calculate molarity of sucrose solution:

Molar mass of sucrose (C₁₂H₂₂O₁₁) = 342.3 g/mol

\(M_{\text{sucrose}} = \frac{5}{342.3} \times \frac{1000}{100} = 0.146 \text{M}\)

Compare osmotic pressures:

Both are non-electrolytes (i = 1), so osmotic pressure is proportional to molarity

Since \(M_{\text{glucose}} > M_{\text{sucrose}}\), glucose solution has higher osmotic pressure

Determine tonicity:

A solution with higher osmotic pressure is hypertonic relative to a solution with lower osmotic pressure

This is important for understanding how different sugar solutions affect cells. Hypertonic solutions cause cell shrinkage, while hypotonic solutions cause cell swelling.

5% glucose solution is hypertonic relative to 5% sucrose solution

10

Cell in Sodium Hydrogen Phosphate Solution

What happens to a living cell (osmotic pressure = 1100 kPa) placed in a 2% sodium hydrogen phosphate solution at 36°C? (Solution density = 1.06 g/mL, i = 2.6)

Step-by-Step Solution:

Convert cell osmotic pressure to atm:

\(\pi_{\text{cell}} = 1100 \text{kPa} \times \frac{1 \text{atm}}{101.325 \text{kPa}} = 10.86 \text{atm}\)

Calculate molar mass of Na₂HPO₄:

\(M = 2(22.99) + 1.01 + 30.97 + 4(16.00) = 141.96 \text{g/mol}\)

Calculate volume of 100 g solution:

\(V = \frac{100 \text{g}}{1.06 \frac{\text{g}}{\text{mL}}} = 94.34 \text{mL} = 0.09434 \text{L}\)

Calculate moles of Na₂HPO₄ in 100 g solution:

\(n = \frac{2}{141.96} = 0.01409 \text{mol}\)

Calculate molarity:

\(M = \frac{0.01409}{0.09434} = 0.1493 \text{M}\)

Calculate osmotic pressure of solution:

\(\pi = i \cdot M \cdot R \cdot T = 2.6 \times 0.1493 \times 0.0821 \times 309\)

\(\pi = 9.85 \text{atm}\)

Compare osmotic pressures:

Cell osmotic pressure = 10.86 atm

Solution osmotic pressure = 9.85 atm

Since \(\pi_{\text{solution}} < \pi_{\text{cell}}\), the solution is hypotonic

In a hypotonic solution, water moves into the cell, causing it to swell. This is critical for understanding how IV solutions affect cells.

The cell will swell (hypotonic solution)

11

Potassium Citrate Solution Preparation

How to prepare 200 mL of potassium citrate solution from crystalline potassium citrate and distilled water if it boils at 100.067°C? (\(K_b = 1.86 ^\circ\text{C·kg/mol}\), i = 3.8, density = 1 g/mL)

Step-by-Step Solution:

Calculate boiling point elevation:

\(\Delta T_b = 100.067 - 100.000 = 0.067 ^\circ\text{C}\)

Apply boiling point elevation formula:

\(\Delta T_b = i \cdot K_b \cdot m\)

Rearrange to solve for molality (m):

\(m = \frac{\Delta T_b}{i \cdot K_b} = \frac{0.067}{3.8 \times 1.86}\)

Calculate molality:

\(m = \frac{0.067}{7.068} = 0.00948 \text{mol/kg}\)

Calculate moles of potassium citrate needed:

Mass of solvent = 200 g (since density = 1 g/mL)

\(n = m \times \text{mass of solvent} = 0.00948 \frac{\text{mol}}{\text{kg}} \times 0.200 \text{kg} = 0.001896 \text{mol}\)

Calculate mass of potassium citrate:

Molar mass of K₃C₆H₅O₇ = 3(39.10) + 6(12.01) + 5(1.01) + 7(16.00) = 306.3 g/mol

\(\text{Mass} = 0.001896 \text{mol} \times 306.3 \frac{\text{g}}{\text{mol}} = 0.581 \text{g}\)

This calculation shows how to prepare a solution with specific colligative properties, which is essential for formulating pharmaceutical solutions with desired characteristics.

Dissolve 0.581 g of potassium citrate in 200 mL of distilled water

12

Mass Percentage of Papaverine Hydrochloride

Calculate the mass percentage concentration of a papaverine hydrochloride solution that freezes at -0.024°C. (\(K_f = 0.52 ^\circ\text{C·kg/mol}\), i = 1.7)

Step-by-Step Solution:

Calculate freezing point depression:

\(\Delta T_f = 0 - (-0.024) = 0.024 ^\circ\text{C}\)

Apply freezing point depression formula:

\(\Delta T_f = i \cdot K_f \cdot m\)

Rearrange to solve for molality (m):

\(m = \frac{\Delta T_f}{i \cdot K_f} = \frac{0.024}{1.7 \times 0.52}\)

Calculate molality:

\(m = \frac{0.024}{0.884} = 0.02715 \text{mol/kg}\)

Calculate mass of solute per kg solvent:

Molar mass of papaverine HCl = 375.5 g/mol

\(\text{Mass} = m \times \text{molar mass} = 0.02715 \frac{\text{mol}}{\text{kg}} \times 375.5 \frac{\text{g}}{\text{mol}} = 10.19 \frac{\text{g}}{\text{kg}}\)

Calculate mass percentage:

\(\text{Mass \%} = \frac{10.19}{1000 + 10.19} \times 100 = 1.01\%\)

This calculation is essential for determining the concentration of injectable medications to ensure proper therapeutic effects without causing tissue irritation.

The mass percentage concentration is 1.01%