Question

If 0.01 mole of NaOH is added to 1 litre solution of a basic buffer having $pH=8.7$ initially, the $pH$ changes to 9. Calculate the buffer capacity of the buffer.

• A. 0.33
• B. 0.033
• C. 0.087
• D. 0.02

Answer 1

The correct option is B 0.033

Finding the change in $pH$ i.e. $\Delta pH$
$\Delta pH=9-8.7=0.3$
Amount of strong base $NaOH$ added :

$\text{Concentration}=\frac{\text{Moles}}{\text{Volume}}=\frac{0.01}{1}=0.01M$

$\text{Buffer Capacity}=\frac{\text{Amount of NaOH added}}{\Delta pH}\text{Buffer Capacity}=\frac{0.01}{0.3}=0.033$


Theory :

Buffer capacity :
It is defined as the moles of a strong acid or strong base required to change the pH of 1 L of a buffer by one unit.

Mathematically, it is the ratio of the small amount of acid or base added to the change in pH caused in the buffer.

Consider two buffer solutions named as buffer 1 and buffer 2 :
Buffer-1
1 L solution of $0.7MC{H}_{3}COONa+0.3MC{H}_{3}COOH$having pH = 5
Buffer-2
1 L solution of $0.07MC{H}_{3}COONa+0.03MC{H}_{3}COOH$ having pH = 5.
On addition to acid externally, concentration of anion of salt decreases whereas concentration of acid increases by the same amount.
On addition of 0.06M HCl in both the buffer solutions, pH changes as follows
Final pH :
Buffer - 1 :
$p{H}_f=4.74+log\frac{0.7-0.06}{0.3+0.06}=4.97$
Buffer - 2 :
$p{H}_f=4.74+log\frac{0.07-0.06}{0.03+0.06}=3.78$
Change in pH :
Buffer - 1:
$\Delta pH=5-4.97=0.03$
Buffer capacity $=\frac{0.06}{0.03}=2$
Buffer - 2:
$\Delta pH=5-3.78=1.22$
Buffer capacity $=\frac{0.06}{1.22}=0.049$