Dissociation constand of propanoic acid is 1.76. Determine degree of dissociation of P ropanoic acid in 0.2molar in aqueous solution?
To determine the degree of dissociation of propanoic acid in a 0.2 molar aqueous solution, you need to use the dissociation constant.
The dissociation constant (Ka) is defined as the ratio of the concentration of the products to the concentration of the reactants for a dissociation reaction. In this case, the dissociation reaction of propanoic acid can be represented as follows:
CH3CH2COOH ⇌ CH3CH2COO- + H+
The dissociation constant expression can be written as:
Ka = [CH3CH2COO-][H+] / [CH3CH2COOH]
We are given that the dissociation constant (Ka) for propanoic acid is 1.76. For a weak acid like propanoic acid, the degree of dissociation (α) is typically much less than 1. The degree of dissociation is defined as the ratio of the concentration of the dissociated species ([CH3CH2COO-] and [H+]) to the initial concentration of the acid ([CH3CH2COOH]).
Let's assume the degree of dissociation of propanoic acid is α.
Since we are given the molar concentration of the propanoic acid solution (0.2 M), we can substitute the given values into the dissociation constant expression:
1.76 = (α * 0.2) * (α * 0.2) / (0.2 - α * 0.2)
Simplifying and rearranging the equation will allow us to solve for α:
1.76 = (α^2 * 0.04) / (0.2 - α * 0.2)
Multiply both sides of the equation by (0.2 - α * 0.2):
1.76 * (0.2 - α * 0.2) = α^2 * 0.04
0.352 - 0.0352α = 0.04α^2
0.04α^2 + 0.0352α - 0.352 = 0
Now, we can solve this quadratic equation for α using a quadratic formula or numerical methods. The solutions obtained will give us the positive degree of dissociation.
Solving the equation, we get that α ≈ 0.247.
Therefore, the degree of dissociation of propanoic acid in a 0.2 molar aqueous solution is approximately 0.247, or 24.7%.