Question

What is the percent dissociation of a 0.55 M solution of weak acid HC3H5O3 (Ka = 1.4 × 10−4)?

Answer 1

• 1.6%

Step-by-step explanation:

Percent dissociation is the amount of the substance dissociated per 100 parts.

Determine how much dissociate from a 0.55 M solution.

1. Write the dissociation equation:

C₃H₅O₃H ⇄ C₃H₅O₃H⁻ + H⁺

2. Build the ICE (initial, change, equilibrium) table:

C₃H₅O₃H ⇄ C₃H₅O₃H⁻ + H⁺

I 0.55M 0 0

C - x +x +x

E 0.55 - x x x

3. Write the equilibrium equation and calculate x

• Keq = x² / (0.55 - x)

• 1.4 × 10⁻⁴ = x² / (0.55 -x)

• x² + 0.00014x - 0.000077 = 0

Use the quadratic equation:

`x=(-0.00014\pm√(0.00014^2-4(1)(-0.000077)))/(2(1))`

The positve value is x = 0.0087

4. Calculate the percent dissociation, α

• α = (0.0087/0.55) × 100

• α = 1.6% ← answer