Consider a weak base of . mL of M HCl and mL of M weak base are mixed to make mL of a buffer of pH at . The values of and respectively are:
[Given: , , ]
- A
,
- B
,
- C
,
- D
,
Consider a weak base of . mL of M HCl and mL of M weak base are mixed to make mL of a buffer of pH at . The values of and respectively are:
[Given: , , ]
,
,
,
,
Correct answer:D
Standard Method
Given: A weak base has . mL of M HCl and mL of M weak base are mixed to form mL buffer of pH .
Find: The values of and , and hence the correct option.
For a basic buffer, first convert into of the conjugate acid .
Now apply the Henderson–Hasselbalch equation for the buffer :
Using ,
Moles of HCl added are:
Moles of base added are:
After neutralisation, all added HCl converts an equal amount of into . Therefore,
So the buffer ratio becomes:
Since both are prepared from solutions of the same molarity, the common factor cancels:
Also, total volume is mL:
Therefore, mL and mL. The correct option is D.
Why the mole ratio becomes volume ratio
Given: Both HCl and weak base have concentration M.
Find: Why can be written directly.
The base buffer contains weak base and its conjugate acid formed by partial neutralisation:
Initial moles:
After reaction:
Hence,
Since the required ratio is ,
This directly gives the same result used in the standard solution.
Using the Henderson–Hasselbalch form for a weak acid without first converting to . This is wrong because the buffer here is written in terms of and pH must be related to of . First use .
Taking equal to the initial moles of base. This is wrong because is produced only by neutralisation with HCl. The correct amount is the moles of HCl added, while remaining base is initial base minus reacted base.
Writing the buffer ratio as instead of . This reverses the ratio and leads to incorrect values of and . Use the equation exactly as obtained from .
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