The pH of an acid buffer can be raised by 2 units by:

Increasing the concentration of both weak acid and salt by two moles

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Increasing the concentration of both the acid and salt by 10 times

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Diluting the solution by 10 times

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Increasing the concentration of the salt by 10 times and decreasing concentration of the acid by 10 times

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D Increasing the concentration of the salt by 10 times and decreasing concentration of the acid by 10 times
For an acidic buffer solution, we use the Henderson-Hasselbalch equation
Or
$p H = p K_{a} + l o g \frac{[s a l t]}{[a c i d]}$
So Case I
$p H_{1} = p K_{a} + l o g \frac{[s a l t_{1}]}{[a c i d_{1}]}$ ....(1)
So Case 2
$p H_{2} = p K_{a} + l o g \frac{[s a l t_{2}]}{[a c i d_{2}]}$ ....(2)
According to question,
(2)-(1) =2; $p K_{a}$ terms will get cancelled we get
2 = $l o g \frac{[a c i d_{1}] \times [s a l t_{2}]}{[s a l t_{1}] \times [a c i d_{2}]}$
taking antilog and solving
$\frac{[a c i d_{1}] \times [s a l t_{2}]}{[s a l t_{1}] \times [a c i d_{2}]} = 10^{2}$
When the concentration of salt is increased by 10 times and concentration of acid is decreased by 10 times, then we get this ratio.

Suggest Corrections