## Part b - Milliequivalents and osmolarity

*A 42 year old male (weight 62kg) requires KCl supplementation due to hypokalaemia. The physician orders for 40 mEq KCl to be given intravenously. For KCl to be infused peripherally, hospital policy recommends 10 mEq/100ml as the maximum concentration and a maximum administration rate of 10 mEq/hr. Your pharmacy stocks KCl 20mEq/50ml solution. *

*i) How would you prepare the KCl solution for peripheral administration to the patient and at what rate would you instruct the nurse to administer the medication? (2 marks)*

Step 1:

Rx is **40 mEq KCl**

The pharmacy keeps 20 mEq/50ml KCl vials, this means each vial has 20 mEq of KCl.

Therefore **two** 20 mEq/50ml vials is required to obtain total of **40 mEq KCl in 100ml**.

Step 2:

Maximum concentration recommended is **10 mEq/100ml**, so dilution is required.

To dilute the 40 mEq/100ml to 10 mEq/100ml add 300ml of diluent (usually normal saline) to make a final concentration of **40 mEq in 400ml**.

Step 3:

The maximum rate allowed is 10 mEq per hour, therefore instructions are administer 10 mEq/100ml per hour for four hours to give the patient a total of 40 mEq KCl.

*ii) What is the total osmolarity ( mOsmol/L) of the above IV solution if 0.9% NaCl was used as the diluent? (KCl = 74.5; NaCl= 58.4) (3 marks)*

Osmolarity is to do with the number of osmotically active particles in a solution. Each osmole is equivalent to the number of **moles** of active particle present. The more osmotically concentrated a solution is the stronger its pull is on the movement on water. Salts such as NaCl have two osmotically active components, Na+ and Cl-, which means one mole of NaCl is equivalent to **two** osmoles.

1 mole of NaCl - > 2 Osmoles NaCl

1 mole of KCl - > 2 Osmoles KCl

**KCl**

A chemical **equivalent** is an old fashioned term meaning how much of an electrolyte (in moles) is required to react with one mole of H+ ions. It essentially works out to be the same/proportional to the valency of the ion in question.

1 mEq of KCl is the same as 1 mmol of KCl.

40 mEq of KCl -> 40 mmol KCl

-> 80 mOsm KCl

80 mOsm KCl in 400 ml

The question requires the answer to be in **mOsm/L**.

To convert 80 mOsm/400ml to how much this is in 1000ml -> 80*(1000/400) = 80*(10/4)

=> **200 mOsm/L **

**NaCl**

As the original stock solution of KCl was 40mEq in 100ml which needs to be diluted to **400 **ml by adding 300 ml of 0.9% NaCl saline, the saline has also been diluted and we need to work out the new concentration for NaCl.

0.9% NaCl:

= 0.9g/100ml= 900mg/100ml = 9000mg/1000ml = 9g/L

300ml of 0.9% NaCl contains 9*(300/1000) = 2.7 g NaCl

To convert 2.7g of NaCl to moles :

MW of NaCl is 58.4

2.7/58.4 = 0.04623 or 46.23 mmol of NaCl

The new concentration of NaCl is 46.23 mmol/**400ml**

With this concentration to find out what the equivalent 1 L solution is multiply by 1000/400 (or 10/4)

46.23*(10/4) = 115.6 mmol/L

1 mole of NaCl is equivalent to 2 Osmoles

115.6*2 -> **231 mOsm/L **

**Total osmolarity of the solution** = 200 + 231 = **431**** mOsm/L**