Question

When a sample of MegaCaff that weighs 0.5 g and has a total surface area of 0.25 m² was dissolved in water for 2 minutes, the dissolution rate constant (k) was found to be 0.01 cm/min. The saturation solubility is known to be 3.2 x 10^-3 g/mL.

If the material is now pressed into a tablet with surface area 2.0 cm², what will be the effect on the dissolution rate (dM/dt) and the dissolution rate constant (k)?

Answer 1

The dissolution rate constant (K)

dM/dt =K *S * (Cs-Ct)

dM/dt = 0.5 g / 2 mintues = 0.25 g/min

S = 0.25 m²

Cs = 3.2 *10^-3 g/mL

K = 0.01 cm/min

1m = 100 cm

0.25 g/min = 0.01 cm/min * (0.25 * 100*100 cm²) (0.0032 g/mL - Ct)

(0.0032 g/mL - Ct) =   0.25 g/min / 25 cm³/min = 0.01 g/cm³

(0.0032 g/mL - 0.01 g/cm³ ) = Ct

(1 mL = 1 cm³)

Ct = -0.0068 g/mL

If the material is now pressed into a tablet with surface area 2.0 cm²

Calculate the dissolution rate constant (K)

dM/dt =K *S * (Cs-Ct)

dM/dt =0.5 g / 2 mintues = 0.25 g/min

S = 2.0 cm²

Cs = 3.2 *10^-3 g/mL, Ct = -0.0068 g/mL

K = 0.01 cm/min

1m = 100 cm

dM/dt = 0.01cm/min* (2 cm²) * (0.0032 g/mL - (-0.0068)g/ mL)

dM/dt = 0.02 cm³/min *0.01 g/mL = 2*10^-4 g/min (1mL = 1cm³)

dM/dt = 2*10^-4 g/min at Surface area = 2 cm²

dM/dt = 0.5 g / 2 mintues = 0.25 g/min at Surface area = 0.25 m²

With the decrease in the Surface area dissolution rate decreases and Dissolution rate constant will be same