1. **Stating the problem:** Given a one-compartment pharmacokinetic model with a 300 mg dose and plasma concentration ($C_p$) data at various times, we need to calculate the elimination rate constant ($k$), initial plasma concentration ($C_p^0$), volume of distribution ($V_D$), half-life ($t_{1/2}$), and total clearance ($Cl_T$). 2. **Calculate $k$ (elimination rate constant):** Using two data points where concentration decreases exponentially, choose times 0.5 hr ($C_p=70.50$) and 3.0 hr ($C_p=21.00$). $$k = \frac{2.3(\log Y_2 - \log Y_1)}{X_2 - X_1} = \frac{2.3(\log 21.00 - \log 70.50)}{3.0 - 0.5}$$ Calculate logs: $\log 21.00 \approx 1.322$, $\log 70.50 \approx 1.848$ So, $$k = \frac{2.3(1.322 - 1.848)}{2.5} = \frac{2.3(-0.526)}{2.5} = \frac{-1.2098}{2.5} = -0.484 \, \text{hr}^{-1}$$ Since $k$ is a rate constant, take the positive value: $$k = 0.484 \, \text{hr}^{-1}$$ 3. **Calculate initial plasma concentration $C_p^0$:** Use the linear equation: $$\ln C_p = -kt + \ln C_p^0$$ At $t=0.5$ hr, $C_p=70.50$: $$\ln 70.50 = -0.484 \times 0.5 + \ln C_p^0$$ Calculate: $\ln 70.50 \approx 4.256$ So, $$4.256 = -0.242 + \ln C_p^0$$ $$\ln C_p^0 = 4.256 + 0.242 = 4.498$$ Therefore, $$C_p^0 = e^{4.498} \approx 89.74 \, \mu g/mL$$ 4. **Calculate volume of distribution $V_D$:** Given dose $D_B^0 = 300$ mg and $C_p^0=89.74$ $\mu g/mL = 0.08974$ mg/mL, $$V_D = \frac{D_B^0}{C_p^0} = \frac{300}{0.08974} \approx 3342.5 \, mL = 3.3425 \, L$$ 5. **Calculate half-life $t_{1/2}$:** $$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.484} \approx 1.43 \, \text{hours}$$ 6. **Calculate total clearance $Cl_T$:** $$Cl_T = k \times V_D = 0.484 \times 3.3425 = 1.618 \, L/hr$$ --- **Final answers:** - $k = 0.484 \, \text{hr}^{-1}$ - $C_p^0 = 89.74 \, \mu g/mL$ - $V_D = 3.34 \, L$ - $t_{1/2} = 1.43 \, \text{hours}$ - $Cl_T = 1.62 \, L/hr$