A patient requires 36 mmol of phosphate and 90 meq of potassium in their PN. The pharmacy has stock solutions of potassium phosphate (3 mmol of phosphate and 4.4 meq potassium per ml) and potassium chloride (2 meq potassium/ml). How many milliliters of potassium chloride are required?

To meet the patient's potassium requirement, 18.6 ml of potassium chloride solution is needed after considering potassium provided by the potassium phosphate solution. The total potassium needs are 90 meq, with 52.8 meq provided by 12 ml of potassium phosphate. The remaining 37.2 meq is met by administering 18.6 ml of potassium chloride.

To determine how many milliliters of potassium chloride (KCl) are needed, we first need to calculate the total potassium (K) requirement not met by the potassium phosphate (K₃PO₄) solution.

Step-by-Step Solution:

The patient needs 90 meq of potassium (K) in total.
The potassium phosphate solution provides both potassium and phosphate, with 3 mmol of phosphate and 4.4 meq of potassium per ml.
We need 36 mmol of phosphate, so we calculate the required volume of potassium phosphate solution:
(36 mmol PO₄ / 3 mmol PO₄ per ml) = 12 ml of K₃PO₄ solution.
This 12 ml of potassium phosphate solution provides:
(12 ml x 4.4 meq K per ml) = 52.8 meq of K.
Subtract the potassium provided by the phosphate solution from the total potassium requirement:
(90 meq K - 52.8 meq K) = 37.2 meq of K required from potassium chloride.
The potassium chloride solution concentration is 2 meq K per ml. Therefore, we calculate the required volume of potassium chloride solution:
(37.2 meq K / 2 meq K per ml) = 18.6 ml of KCl solution.

Thus, 18.6 ml of potassium chloride are required to meet the patient’s potassium needs.