Question
1. A patient that is administered 500mg of oral procainamide as a capsule. It's known from prior clinic visit that the patient has a half life equal to 4 hr, and a volume of distribution of 175 l.
The capsule that's administered to the patient has an absorption rate constant equal to 2 hr-1, and an oral bioavailability friction of 0.85.
Calculate the procainamide serum concentration 4 hr after a single would be equal to:
2. A patient receiving 24 meq of lithium carbonate capsule. From previous clinic visits , it's known that the patient has a volume of distribution of 69l and an elimination rate constant equal to 0.9.
Calculate the serum lithium concentration 12 hr after a single dose?
The capsule that's administered to the patient has an absorption rate constant equal to 2 hr-1, and an oral bioavailability friction of 0.85.
Calculate the procainamide serum concentration 4 hr after a single would be equal to:
2. A patient receiving 24 meq of lithium carbonate capsule. From previous clinic visits , it's known that the patient has a volume of distribution of 69l and an elimination rate constant equal to 0.9.
Calculate the serum lithium concentration 12 hr after a single dose?
Answers
Answered by
GPT-4o mini
To calculate the serum concentration of both procainamide and lithium at the specified times, we will use the respective pharmacokinetic equations for each drug.
### 1. Procainamide Calculation
**Given:**
- Dose (D) = 500 mg
- Half-life (\(t_{1/2}\)) = 4 hours
- Volume of distribution (Vd) = 175 L
- Absorption rate constant (\(k_a\)) = 2 hr\(^{-1}\)
- Bioavailability (F) = 0.85
- Time (t) = 4 hours
**First, we need to determine the elimination rate constant (\(k\)):**
\[
k = \frac{0.693}{t_{1/2}} = \frac{0.693}{4} = 0.1733 \text{ hr}^{-1}
\]
**Next, we calculate the expected concentration at time t using the following equation:**
The concentration after a single oral dose can be calculated as follows:
\[
C(t) = \frac{F \cdot D \cdot k_a}{V_d \cdot (k_a - k)} \cdot \left( e^{-k \cdot t} - e^{-k_a \cdot t} \right)
\]
Substituting the values:
\[
C(4) = \frac{0.85 \cdot 500 \cdot 2}{175 \cdot (2 - 0.1733)} \cdot \left( e^{-0.1733 \cdot 4} - e^{-2 \cdot 4} \right)
\]
Calculating each component step:
1. Calculate \(k_a - k\):
\[
k_a - k = 2 - 0.1733 = 1.8267
\]
2. Calculate \(e^{-0.1733 \cdot 4}\):
\[
e^{-0.1733 \cdot 4} = e^{-0.6932} \approx 0.5
\]
3. Calculate \(e^{-2 \cdot 4}\):
\[
e^{-8} \approx 0.000335
\]
4. Substitute back:
\[
C(4) = \frac{0.85 \cdot 500 \cdot 2}{175 \cdot 1.8267} \cdot (0.5 - 0.000335)
\]
\[
C(4) = \frac{850}{319.67} \cdot (0.499665)
\]
\[
C(4) = 2.66 \cdot 0.499665 \approx 1.33 \text{ mg/L}
\]
So the procainamide serum concentration 4 hours after the dose would be approximately **1.33 mg/L**.
### 2. Lithium Calculation
**Given:**
- Dose (D) = 24 mEq (Note: 1 mEq of lithium is approximately 0.3 g of lithium carbonate, so 24 mEq corresponds to about 7.2 g of lithium carbonate, but we can proceed with mEq)
- Volume of distribution (Vd) = 69 L
- Elimination rate constant (k) = 0.9 hr\(^{-1}\)
- Time (t) = 12 hours
**We can directly calculate serum concentrations using:**
\[
C(t) = \frac{D}{V_d} \cdot e^{-k \cdot t}
\]
Substituting the values:
\[
C(12) = \frac{24}{69} \cdot e^{-0.9 \cdot 12}
\]
Calculating the exponent:
\[
e^{-10.8} \approx 0.000204
\]
Substituting back:
\[
C(12) = \frac{24}{69} \cdot 0.000204
\]
Calculating:
\[
C(12) = 0.3478 \cdot 0.000204 \approx 0.000071 \text{ mEq/L}
\]
So the serum lithium concentration 12 hours after a single dose would be approximately **0.000071 mEq/L**.
Ensure that the value makes clinical sense in the context of therapeutic ranges for lithium.
### 1. Procainamide Calculation
**Given:**
- Dose (D) = 500 mg
- Half-life (\(t_{1/2}\)) = 4 hours
- Volume of distribution (Vd) = 175 L
- Absorption rate constant (\(k_a\)) = 2 hr\(^{-1}\)
- Bioavailability (F) = 0.85
- Time (t) = 4 hours
**First, we need to determine the elimination rate constant (\(k\)):**
\[
k = \frac{0.693}{t_{1/2}} = \frac{0.693}{4} = 0.1733 \text{ hr}^{-1}
\]
**Next, we calculate the expected concentration at time t using the following equation:**
The concentration after a single oral dose can be calculated as follows:
\[
C(t) = \frac{F \cdot D \cdot k_a}{V_d \cdot (k_a - k)} \cdot \left( e^{-k \cdot t} - e^{-k_a \cdot t} \right)
\]
Substituting the values:
\[
C(4) = \frac{0.85 \cdot 500 \cdot 2}{175 \cdot (2 - 0.1733)} \cdot \left( e^{-0.1733 \cdot 4} - e^{-2 \cdot 4} \right)
\]
Calculating each component step:
1. Calculate \(k_a - k\):
\[
k_a - k = 2 - 0.1733 = 1.8267
\]
2. Calculate \(e^{-0.1733 \cdot 4}\):
\[
e^{-0.1733 \cdot 4} = e^{-0.6932} \approx 0.5
\]
3. Calculate \(e^{-2 \cdot 4}\):
\[
e^{-8} \approx 0.000335
\]
4. Substitute back:
\[
C(4) = \frac{0.85 \cdot 500 \cdot 2}{175 \cdot 1.8267} \cdot (0.5 - 0.000335)
\]
\[
C(4) = \frac{850}{319.67} \cdot (0.499665)
\]
\[
C(4) = 2.66 \cdot 0.499665 \approx 1.33 \text{ mg/L}
\]
So the procainamide serum concentration 4 hours after the dose would be approximately **1.33 mg/L**.
### 2. Lithium Calculation
**Given:**
- Dose (D) = 24 mEq (Note: 1 mEq of lithium is approximately 0.3 g of lithium carbonate, so 24 mEq corresponds to about 7.2 g of lithium carbonate, but we can proceed with mEq)
- Volume of distribution (Vd) = 69 L
- Elimination rate constant (k) = 0.9 hr\(^{-1}\)
- Time (t) = 12 hours
**We can directly calculate serum concentrations using:**
\[
C(t) = \frac{D}{V_d} \cdot e^{-k \cdot t}
\]
Substituting the values:
\[
C(12) = \frac{24}{69} \cdot e^{-0.9 \cdot 12}
\]
Calculating the exponent:
\[
e^{-10.8} \approx 0.000204
\]
Substituting back:
\[
C(12) = \frac{24}{69} \cdot 0.000204
\]
Calculating:
\[
C(12) = 0.3478 \cdot 0.000204 \approx 0.000071 \text{ mEq/L}
\]
So the serum lithium concentration 12 hours after a single dose would be approximately **0.000071 mEq/L**.
Ensure that the value makes clinical sense in the context of therapeutic ranges for lithium.