Ternary Phase Diagram Notes

Ternary Phase Diagram

Basics For Triphasic Chart

• At each apex of the triangle, the concentration of one component is $100\%w/w$.

• At any point inside the triangle, the sum of the concentrations of the three components is $100\%$.

• Any point inside the curve represents a two-phase system, while points outside the curve represent a one-phase system.

Problems (1)

Plotting points on a triangular coordinate graph paper:

• Several points are provided with varying percentages of components A, B, and C.

• For example:

  • A: $30\%$, B: $30\%$, C: $40\%$
  • A: $84\%$, B: $11\%$, C: $5\%$
  • A: $78\%$, B: $12\%$, C: $10\%$

Analysis from the Graph

1. Solubility:

   • a. the solubility of component A in component B
   • b. the solubility of component B in component A
   • c. the solubility of component C in component A

2. Formulation Scenario:

   • A pharmacist needs to formulate a single-phase solution with $50\% w/w$ of component B and $35\%$ of component C.

   • This solution is to be diluted with component A immediately before use.

   • The question asks for the sequence of phase changes observed as component A is progressively added.

Determining Solubility from the Graph

• Solubility of B in A is $11\%$.

Dilution by Component A

• Phase Change:

  • From 15\%$ A to 31\%$ A: one phase
  • From 31\%$ A to 80\%$ A: two phases
  • From 80\%$ A to 100\%$ A: one phase

Problems (2)

The phase equilibria of a 3-component system (acetic acid, chloroform, and water) was determined.

1. Plot the results on a triangular phase diagram.

2. From the curve, estimate:

   • Weights of the 3 components present in $80$ gm of the one-phase system at the point $(43, 45, 12)$.
   • Weights of the two phases and weights of the 3 components in each phase for $80$ gm of the system having the total composition at the point $(15, 70, 15)$, which is the equilibrium state of the 2 points $(10, 88, 2)$ and $(32, 4, 64)$.

Calculations for Problem 2

1. Weights of components at the point $(43, 45, 12)$ in $80$ gm system:

   • Acetic acid: $80 \times (43/100) = 34.4$ g
   • Chloroform: $80 \times (45/100) = 36$ g
   • Water: $80 - (34.4 + 36) = 9.6$ g

2. Two-Phase System Calculation

   • Using tie line to determine the composition of each phase

   • Lever rule is applied to determine the relative amounts of each phase

   • $X$ represents the overall composition (15, 70, 15)

   • $Y$ and $Z$ represent the two phases in equilibrium (10, 88, 2) and (32, 4, 64) respectively.

   • Lengths of tie lines are used to calculate phase weights

   • $NZ = 10.7$ cm, $NY = 2.9$ cm, $YZ = 13.6$ cm

   • Weight of chloroform rich phase: $(10.7 / 13.6) \times 80 = 62.9$ grams

   • Weight of water rich phase = $80-63=17$ grams

   • or water rich phase: $(2.9 / 13.6) \times 80=17$ grams

3. Component weights in each phase:

   • In Water Rich Phase (W.R.P) (32,4,64)

     • Weight of acetic acid in W.R.P: $(32 / 100) \times 17 = 5.44$ grams
     • Weight of chloroform in W.R.P: $(4 / 100) \times 17 = 0.68$ grams
     • Weight of water in W.R.P: $(64 / 100) \times 17 = 10.88$ grams

   • In Chloroform Rich Phase (Ch.R.P) (10,88,2)

     • Weight of acetic acid in Ch.R.P: $(10 / 100) \times 63 = 6.3$ grams
     • Weight of chloroform in Ch.R.P: $(88 / 100) \times 63 = 55.44$ grams
     • Weight of water in Ch.R.P: $(2 / 100) \times 63 = 1.26$ grams

Problems (3)

• Solubility of component A in component B.

• Solubility of component B in component A.

• Which component is the cosolvent?