The percent dissociation of a 0.29 M solution is
0.210%.
What is buffer solution?
Given that
ka = 1.3 x 10⁻⁵
Initial concentration of HC₃H₅O₂ = 0.29
ka = [H⁺]x[C₃H₅O⁻]/[HC₃H₅O₂]
Let denote the concentration of HC₃H₅O₂ that dissociates to be y.
Therefore,
1.3 x 10⁻⁵ = y * y/0.29 - y
0.000013 = y²/0.29 - y
0.000013(0.29 - y) = y²
0.00000377 - 0.000013y = y²
y² + 0.000013y - 0.00000377 = 0
Let solve
y = (+-b√(b² - 4ac))/2a
where
a = 1
b = 0.000013
c = - 0.00000377
Substitute into y = (+-b√(b² - 4ac))/2a
y = 0.000013 +-√(0.000013)² + 4*1*0.00000377))/2*1
y = 0.000013 +-√(0.0000151)/2
= (0.000013 +- 0.00123)/2
Since concentration cannot be negative
y = (0.000013 + 0.00123)/2
= 0.00609
The concentration of the acide that dissociates is
0.00609
The percentage = (0.00609 * 100)/0.29
= 0.210%.