Name: Basics of Solubility Product
Uploaded: 2019-07-19
Duration: 8 min 38 s
Description: The solubility product constant is a simplified equilibrium constant showing the degree to which a compound can dissociate in water. Let's watch the video.

Question

Calculate the degree of ionization of $0.05 M$ acetic acid if its ${pK}_{a}$ value is $4.74$ . How is the degree of dissociation affected when its solution also contains $0.01 M$ in $HCl$ ?

Accepted Answer

The concentration of acetic acid $C = 0.05 M$ .

${pK}_{a} = 4.74$

${pK}_{a} = - \log [K_{a}]$

$K_{a} = 1.82 \times 10^{- 5}$

$K_{a} = {cα}^{2}$

$α = \sqrt{\frac{K_{a}}{c}} = \sqrt{\frac{1.82 \times 10^{- 5}}{0.05}} = 1.908 \times 10^{- 2} $

When $HCl$ is added to the solution, the concentration of hydrogen ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.

$0.01 M HCl$ is taken.

Let $x$ be the amount of acetic acid dissociated after the addition of $HCl$ .

$Reaction {CH}_{3} COOH ⇌ {CH}_{3} {COO}^{-} + H^{+} Initial concentration 0.05 0 0 after dissociation 0.05 - x x 0.01 + x$

As the dissociation of a very small amount of acetic acid will take place, the values i.e., $0.05 - x$ and $0.01 + x$ can be taken as $0.05 and 0.01$ respectively.

$Ka = \frac{[{CH}_{3} {COO}^{-}] [H^{+}]}{[{CH}_{3} COOH]}$

$x = \frac{1.82 \times 10^{- 5} \times 0.05}{0.01} = 1.82 \times 5 \times 10^{- 5}$

Now,

$α = \frac{Amount of acid dissociation}{Amount of acid taken} = \frac{1.82 \times 5 \times 10^{- 5}}{0.05} = 1.82 \times 10^{- 3}$

$α = 1.82 \times 10^{- 3}$ .

Calculate the degree of ionization of $0.05 M$ acetic acid if its ${pK}_{a}$ value is $4.74$ . How is the degree of dissociation affected when its solution also contains $0.01 M$ in $HCl$ ?

Important Questions on Equilibrium

Learn and Prepare for any exam you want !

Calculate the degree of ionization of 0.05 M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains 0.01 M in HCl?

Important Questions on Equilibrium

Learn and Prepare for any exam you want !

Calculate the degree of ionization of $0.05 M$ acetic acid if its ${pK}_{a}$ value is $4.74$ . How is the degree of dissociation affected when its solution also contains $0.01 M$ in $HCl$ ?