A certain weak acid, HA, has a Ka value of 1.1 * 10-7.

Question:

A certain weak acid, HA, has a Ka value of 1.1 * 10-7.

Part A

Calculate the percent ionization of HA in a 0.10 M solution.

Part B

Calculate the percent ionization of HA in a 0.010 M solution.

Calculating the Degree of Dissociation of Weak Acids

Answer:

Consider a powerless monoprotic corrosive of the structure HA . Give its focus a chance to be c and separation steady be KaConsider a frail monoprotic corrosive of the structure HA . Give its fixation a chance to be c and separation steady be Ka

The general concoction condition delineating the ionization of HA is:

HA ⇋ H + A−

c(1−α) cα

α is the level of separation of the acidα is the level of separation of the corrosive

∴Ka=[H+][OH−]/[HA]

⇒Ka=(cα)(cα)/c(1−α)

⇒Ka=cα2/1−α(1)

⇒cα2+αKa−Ka=0

Henceforth, the level of separation can be assessed by illuminating the above quadratic condition.

At the point when the separation consistent Ka has an extremely little esteem, we can utilize approximations in condition (1) to discover the level of separation.

1−α≈1

⇒Ka=cα2

⇒α=√Ka/c

Answer and Explanation:

(a)Ka=1.1×10−7

c=0.1 M

∴α=√Ka/c

⇒α=√1.1×10−7/0.1

⇒α=1.048×10−3=0.1048%

(b)Ka=1.1×10−7

c=0.01 M

∴α=√Ka/c

⇒α=√1.1×10−7/0.01

⇒α=3.316×10−3=0.3136%

We can without much of a stretch see that on weakening from 0.1 M to 0.01 M, the level of separation of HA has expanded.