Question

If the ionization (dissociation) constant of acetic acid is ${ k }_{ a }$, what will be the pH of a solution containing equal concentrations of acetic acid and sodium acetate?

Answer 1

Hint: Henderson - Hasselbalch equation for a weak acid is given by :
${ pH }={ p }{ k }_{ a }+log\dfrac { \left[ salt \right] }{ \left[ acid \right] } $.
By substituting the conditions given in the question, we will get the required answer. The condition given here is that the concentration of acetic acid and sodium acetate are equal, which means that the term $\dfrac { \left[ salt \right] }{ \left[ acid \right] } $ will be equal to 1 . log 1 = 0.

Complete step by step answer:
We have been provided with the ionization constant of acetic acid as ka.
Also, it has been given that both acetic acid $\left( { CH }_{ 3 }{ COOH } \right) $ and sodium acetate $\left( { CH }_{ 3 }{ COONa } \right) $ have the same concentration.
i.e,
$\left[ { CH }_{ 3 }{ COOH } \right] =\left[ { CH }_{ 3 }{ COONa } \right] $
We need to determine the pH of the solution.

For that we can use the Henderson-Hasselbalch equation, which is given by :
${ pH }={ p }{ k }_{ a }+log\dfrac { \left[ salt \right] }{ \left[ acid \right] } $
${ pH }={ p }{ k }_{ a }+log\dfrac { \left[ { CH }_{ 3 }{ COONa } \right] }{ \left[ { CH }_{ 3 }{ COOH } \right] }$
And since it is already given that
$\left[ { CH }_{ 3 }{ COOH } \right] =\left[ { CH }_{ 3 }{ COONa } \right] $
pH = ${pk }_{ a }$ + log 1
We know that log 1 = 0.
pH =${pk }_{ a }$.
And hence, the answer is pH =${ pk }_{ a }$.

Additional Information:
A strong acid is the one which can ionize completely in water, whereas a weak acid is that which can dissociate completely in a particular medium.
Ethanoic acid is a weak acid. When combined with water it produces hydronium ions and ethanoate ions, but the backward reaction is more successful than the forward one.
At one time, only about 1% of the ethanoic acid molecules have converted into ethanoate ions. The rest remain as such as simple ethanoic acid molecules.
Most of the organic acids are weak. Hydrogen fluoride (in combination with water produces hydrofluoric acid) is a weak inorganic acid that you may come across anywhere.
The dissociation constant of ${ CH }_{ 3 }{ COOH }$ is given by:
${ CH }_{ 3 }{ COOH }+{ H }_{ 2 }{ O }\rightarrow { CH }_{ 3 }{ CO{ O }^{ - } }+{ H }_{ 3 }{ O }^{ + }$
Let us consider at time , t=0 :
Concentration of ${ CH }_{ 3 }{ COOH }$ is ‘c’
Concentration of ${ CH }_{ 3 }{ CO{ O }^{ - } }$ and H+ are 0.
At a particular time t =${ t }_{ 1 }$ ,
Concentration of ${ CH }_{ 3 }{ COOH }$ is ${ 'c-c\alpha ' }$
Concentration of ${ CH }_{ 3 }{ CO{ O }^{ - } }$ and ${ H }^{ + }$ are $c\alpha $ and $c\alpha $ respectively.

Now , at equilibrium , the equilibrium constant for the above reaction is written as
${ K }_{ C }=\dfrac { \left[ { CH }_{ 3 }{ COO }^{ - } \right] \left[ { H }^{ + } \right] }{ \left[ { CH }_{ 3 }{ COOH } \right] \left[ { H }_{ 2 }{ O } \right] } $
Since the concentration of water is a constant here, we can multiply it with ${ K }_{ C }$ , to get the acid dissociation constant ,${ k }_{ a }$.

${ K }_{ C }\left[ { H }_{ 2 }{ O } \right] ={ k }_{ a }$
${ k }_{ a }=\dfrac { { c\alpha }{ \times c\alpha } }{ c-c\alpha } $
${ k }_{ a }=\dfrac { c{ \alpha }^{ 2 } }{ 1-\alpha } $

Note: From the expression of ${ k }_{ a }$, we can get the value of pka . And again from pka, we can get the value of pH, by the equation,${ pH }={ p }{ k }_{ a }+log\dfrac { \left[ salt \right] }{ \left[ acid \right] } $. In the case of acid - base indicators, for finding their pH range , the same expression is written as pH $={ pK }_{ In }\pm 1$ .