Answer :
To solve this problem, we need to figure out the quantities of sodium chloride and sodium acetate solutions that the pharmacist should add to a parenteral nutrition (PN) solution to meet the specified requirements of 80 mEq of sodium and 45 mEq of acetate. Here is a step-by-step breakdown of the solution:
### Given Values:
- Sodium needed: 80 mEq
- Acetate needed: 45 mEq
### Pharmacy Stock Solutions:
1. Sodium chloride solution contains 4 mEq/mL of sodium.
2. Sodium acetate solution contains 2 mEq/mL of sodium and acetate per mL.
### Step 1: Calculate the volume of sodium acetate needed
Since acetate can only be obtained from sodium acetate (which provides 2 mEq of acetate per mL), we first find out how many milliliters of sodium acetate solution are needed to provide the required 45 mEq of acetate.
[tex]\[ \text{Volume of Sodium Acetate Solution} = \frac{\text{Acetate needed}}{\text{Concentration of acetate in sodium acetate solution}} \][/tex]
[tex]\[ \text{Volume of Sodium Acetate Solution} = \frac{45 \text{ mEq}}{2 \text{ mEq/mL}} = 22.5 \text{ mL} \][/tex]
### Step 2: Calculate the amount of sodium provided by sodium acetate
Next, we determine how much sodium this volume of sodium acetate solution provides. Each milliliter of sodium acetate solution provides 2 mEq of sodium.
[tex]\[ \text{Sodium from Sodium Acetate} = \text{Volume of Sodium Acetate Solution} \times \text{Concentration of sodium in sodium acetate solution} \][/tex]
[tex]\[ \text{Sodium from Sodium Acetate} = 22.5 \text{ mL} \times 2 \text{ mEq/mL} = 45 \text{ mEq} \][/tex]
### Step 3: Calculate the remaining sodium needed from sodium chloride
The total sodium needed is 80 mEq, but we already get 45 mEq from the sodium acetate. Therefore, we need to find out how much more sodium we need from the sodium chloride solution.
[tex]\[ \text{Sodium from Sodium Chloride} = \text{Total Sodium Needed} - \text{Sodium from Sodium Acetate} \][/tex]
[tex]\[ \text{Sodium from Sodium Chloride} = 80 \text{ mEq} - 45 \text{ mEq} = 35 \text{ mEq} \][/tex]
### Step 4: Calculate the volume of sodium chloride needed
Now, we can find the volume of sodium chloride solution required to provide the remaining 35 mEq of sodium. The sodium chloride solution provides 4 mEq of sodium per mL.
[tex]\[ \text{Volume of Sodium Chloride Solution} = \frac{\text{Sodium from Sodium Chloride}}{\text{Concentration of sodium in sodium chloride solution}} \][/tex]
[tex]\[ \text{Volume of Sodium Chloride Solution} = \frac{35 \text{ mEq}}{4 \text{ mEq/mL}} = 8.75 \text{ mL} \][/tex]
### Summary of Required Volumes:
- Volume of Sodium Acetate Solution: 22.5 mL
- Volume of Sodium Chloride Solution: 8.75 mL
Thus, to meet the PN requirements, the pharmacist needs to add:
- 22.5 mL of sodium acetate solution.
- 8.75 mL of sodium chloride solution.
These volumes will ensure that the PN contains exactly 80 mEq of sodium and 45 mEq of acetate.
### Given Values:
- Sodium needed: 80 mEq
- Acetate needed: 45 mEq
### Pharmacy Stock Solutions:
1. Sodium chloride solution contains 4 mEq/mL of sodium.
2. Sodium acetate solution contains 2 mEq/mL of sodium and acetate per mL.
### Step 1: Calculate the volume of sodium acetate needed
Since acetate can only be obtained from sodium acetate (which provides 2 mEq of acetate per mL), we first find out how many milliliters of sodium acetate solution are needed to provide the required 45 mEq of acetate.
[tex]\[ \text{Volume of Sodium Acetate Solution} = \frac{\text{Acetate needed}}{\text{Concentration of acetate in sodium acetate solution}} \][/tex]
[tex]\[ \text{Volume of Sodium Acetate Solution} = \frac{45 \text{ mEq}}{2 \text{ mEq/mL}} = 22.5 \text{ mL} \][/tex]
### Step 2: Calculate the amount of sodium provided by sodium acetate
Next, we determine how much sodium this volume of sodium acetate solution provides. Each milliliter of sodium acetate solution provides 2 mEq of sodium.
[tex]\[ \text{Sodium from Sodium Acetate} = \text{Volume of Sodium Acetate Solution} \times \text{Concentration of sodium in sodium acetate solution} \][/tex]
[tex]\[ \text{Sodium from Sodium Acetate} = 22.5 \text{ mL} \times 2 \text{ mEq/mL} = 45 \text{ mEq} \][/tex]
### Step 3: Calculate the remaining sodium needed from sodium chloride
The total sodium needed is 80 mEq, but we already get 45 mEq from the sodium acetate. Therefore, we need to find out how much more sodium we need from the sodium chloride solution.
[tex]\[ \text{Sodium from Sodium Chloride} = \text{Total Sodium Needed} - \text{Sodium from Sodium Acetate} \][/tex]
[tex]\[ \text{Sodium from Sodium Chloride} = 80 \text{ mEq} - 45 \text{ mEq} = 35 \text{ mEq} \][/tex]
### Step 4: Calculate the volume of sodium chloride needed
Now, we can find the volume of sodium chloride solution required to provide the remaining 35 mEq of sodium. The sodium chloride solution provides 4 mEq of sodium per mL.
[tex]\[ \text{Volume of Sodium Chloride Solution} = \frac{\text{Sodium from Sodium Chloride}}{\text{Concentration of sodium in sodium chloride solution}} \][/tex]
[tex]\[ \text{Volume of Sodium Chloride Solution} = \frac{35 \text{ mEq}}{4 \text{ mEq/mL}} = 8.75 \text{ mL} \][/tex]
### Summary of Required Volumes:
- Volume of Sodium Acetate Solution: 22.5 mL
- Volume of Sodium Chloride Solution: 8.75 mL
Thus, to meet the PN requirements, the pharmacist needs to add:
- 22.5 mL of sodium acetate solution.
- 8.75 mL of sodium chloride solution.
These volumes will ensure that the PN contains exactly 80 mEq of sodium and 45 mEq of acetate.
