Question

20ml of 0.2M sodium hydroxide is added to 50ml of 0.2M acetic acid to give 70ml of the solution. What is the pH of the solution? Calculate the additional volume of 0.2M NaOH required to make the pH of solution 4.74. The ionisation constant of acetic acid is $1.8 \times {10^{ - 5}}$.

Answer 1

Hint: The solution formed by mixing the given amount of acetic acid and NaOH will be an acidic buffer solution. First find out the concentration of the reactants and the products, and then find the pH of the solution. The concentration of added NaOH in the acidic buffer will be equal to the amount of increase and decrease in the concentration of salt and acid. So, now find the new concentrations and equate it to the new pH desired to obtain.

-Number of millimoles = Molarity × Volume (ml) (1)
-pH of any acidic buffer:
 $pH = p{K_a} + \log \dfrac{{\left[ {salt} \right]}}{{\left[ {acid} \right]}}$ (2)

Complete Solution :
-When we add 20ml of 0.2M sodium hydroxide is added to 50ml of 0.2M acetic acid to give a 70ml of solution, an acidic buffer is formed. This reaction can be written as:
$NaOH + C{H_3}COOH \to C{H_3}COONa + {H_2}O$

-From the given values in the question we can calculate the millimoles of NaOH and $C{H_3}COOH$ using equation (1):
Number of millimoles = molarity (M) × volume (ml)
So, millimoles of NaOH involved = M × V = 0.2 × 20 = 4 millimoles
Milimoles of $C{H_3}COOH$ = M × V = 0.2 × 50 = 10 milimoles

But we know that NaOH is a strong base and $C{H_3}COOH$ is a weak acid, so 4 millimoles of NaOH will completely neutralise only 4 millimoles of $C{H_3}COOH$. 6 millimoles of $C{H_3}COOH$ will be left behind and 4 millimoles of $C{H_3}COONa$ and 4 millimoles of ${H_2}O$ will be formed. This can be simplified and written as:
$NaOH + C{H_3}COOH \to C{H_3}COONa + {H_2}O$

Reactants and products	$NaOH$	$C{H_3}COOH$	$C{H_3}COONa$	${H_2}O$
Conc. at t = 0	4	10	0	0
Conc. at t	4 – 4 = 0	10 – 6 = 4	4	4

-So, what is left in the solution is 6 millimoles of $C{H_3}COOH$ which is a weak acid and 4 millimoles of $C{H_3}COONa$ which is a salt of weak acid and strong base. This forms an acidic buffer.
For an acidic buffer the formula to calculate pH is: (equation 2)
$pH = p{K_a} + \log \dfrac{{\left[ {salt} \right]}}{{\left[ {acid} \right]}}$
$pH = p{K_a} + \log \dfrac{{\left[ {C{H_3}COONa} \right]}}{{\left[ {C{H_3}COOH} \right]}}$ (3)
The value of ${K_a}$ is given in the question = $1.8 \times {10^{ - 5}}$.
We know that: $p{K_a} = - \log {K_a}$
We just found out the concentrations of $C{H_3}COOH$ to be 6 millimoles and of $C{H_3}COONa$ to be 4 millimoles. We will put these values in the above equation.
$pH = - \log (1.8 \times {10^{ - 5}}) + \log \dfrac{4}{6}$
 = $4.74 + \log \dfrac{2}{3}$
 = 4.56

-Now according to the question we are adding some additional amount of NaOH to make its pH to 4.74. If a base is dissolved in an acidic buffer then the concentration of acid decreases and the concentration of salt increases. This increase and decrease is equal to the amount of strong base added.
For now let the volume of NaOH added be V (ml).
So the millimoles of NaOH added will be = M × V = 0.2 × V = 0.2V
Amount of acid ($C{H_3}COOH$) was decreased, new concentration of $C{H_3}COOH$= 6 – 0.2V
Amount of salt ($C{H_3}COONa$) was increased, new conc. of $C{H_3}COONa$ = 4 + 0.2V
New pH as given in the question = 4.74
Using equation (3):
$pH = p{K_a} + \log \dfrac{{\left[ {C{H_3}COONa} \right]}}{{\left[ {C{H_3}COOH} \right]}}$
$4.74 = - \log (1.8 \times {10^{ - 5}}) + \log \dfrac{{4 + 0.2V}}{{6 - 0.2V}}$
$4.74 = 4.74 + \log \dfrac{{4 + 0.2V}}{{6 - 0.2V}}$
0 = $\log \dfrac{{4 + 0.2V}}{{6 - 0.2V}}$
We know that log 1 = 0, so:
1 = $\dfrac{{4 + 0.2V}}{{6 - 0.2V}}$
$6 - 0.2V = 4 + 0.2V$
6 – 4 = 0.2V + 0.2V
2 = 0.4V
V = 2 / 0.4
V = 5 ml
So, the volume of NaOH to be added to make the pH of the solution upto 4.74 will be: 5 ml.

Note: On adding a strong base to the acidic buffer the amount of acid decreases and salt increases because: there is decrease in the hydrogen ion concentration by less than the quantity of base added. The hydroxide ions are consumed by the acid and its conjugate base and the pH does not rise much, which should have happened if the buffer system was not being used. Thus, on the basis of Le Chatelier's Principle the equilibrium moves towards right to compensate for loss of hydrogen ions, thus increasing the concentration of salt.