Mass of potassium

Question

A patient requires an IV infusion of an isotonic saline solution containing equal molar amounts of potassium chloride and magnesium chloride. In clinical practice, a saline solution is said to be isotonic if it has the same osmotic pressure as blood plasma (7.74 atm at 36 °C). Give the mass of potassium chloride plus the mass of magnesium chloride in grams that needs to be dissolved in 1.42 L of distilled water to make it isotonic. Assume that the change in volume of the solution after adding the salts can be neglected.

Answer 1

Answer: 6.46 g KCl + 8.25 g MgCl2 = 14.7 g of salt in total.

This is the van't Hoff form of osmotic pressure, and the only real trap in it is counting ions rather than formula units.

1. Total particle concentration. Isotonic means the same osmotic pressure as plasma, so set pi = MRT, where M is the concentration of all dissolved particles:

M = pi / (RT) = 7.74 atm / [(0.08206 L atm per mol K)(309.15 K)] = 0.3051 mol/L

T must be 36 + 273.15 = 309.15 K. Leaving the temperature in Celsius is the most common way this one goes wrong.

2. Moles of particles in the bag. (0.3051 mol/L)(1.42 L) = 0.4332 mol of dissolved particles.

3. The trap. Osmotic pressure is colligative, so it counts particles, not formulas. KCl gives 2 ions per formula unit (K+ and Cl-). MgCl2 gives 3 (Mg2+ plus two Cl-). The problem says equal molar amounts of the two salts, so let n be the moles of each:

2n + 3n = 5n = 0.4332 mol, so n = 0.08665 mol of each salt.

4. Convert to grams.

KCl: (0.08665 mol)(74.55 g/mol) = 6.46 g

MgCl2: (0.08665 mol)(95.21 g/mol) = 8.25 g

Total = 14.7 g

Sanity check worth making a habit: if you had treated each salt as a single particle, you would have called for 2.5 times as much salt as the patient should get. That is the whole reason the van't Hoff factor is the point of this problem. And the line about neglecting the volume change is simply permission to use 1.42 L as the final solution volume instead of solving for it.