What would be the amount (in g) of Span 80 in 5g of a surfactant system composed of a mixture of Span 80 (HLB=4.3) and Tween 80 (HLB=15) to produce a total HLB value of 13?

Steps to Solve

1. Identify given values

We know:

• Total mass of the surfactant system (m_total) = 5 g
• HLB of Span 80 (HLB_A) = 4.3
• HLB of Tween 80 (HLB_B) = 15
• Desired total HLB (HLB_total) = 13

2. Set up the equation for HLB

The equation for the total HLB of a mixture is given by:

$$ HLB_{total} = \frac{(m_A \cdot HLB_A + m_B \cdot HLB_B)}{m_{total}} $$

Where:

• ( m_A ) is the mass of Span 80
• ( m_B ) is the mass of Tween 80

Since ( m_A + m_B = m_{total} = 5 ), we can write ( m_B = 5 - m_A ).

3. Substituting values into the HLB equation

Substituting ( m_B ) into the HLB equation, we get:

$$ 13 = \frac{(m_A \cdot 4.3 + (5 - m_A) \cdot 15)}{5} $$

4. Multiply through by the total mass

Multiply both sides by 5:

$$ 65 = m_A \cdot 4.3 + (5 - m_A) \cdot 15 $$

5. Distribute and rearrange the equation

Distributing gives:

$$ 65 = m_A \cdot 4.3 + 75 - m_A \cdot 15 $$

Rearranging gives:

$$ 65 - 75 = m_A \cdot (4.3 - 15) $$

6. Simplify to find ( m_A )

This simplifies to:

$$ -10 = m_A \cdot (-10.7) $$

Thus,

$$ m_A = \frac{-10}{-10.7} \approx 0.93 \text{ g} $$