Question

PRACTICE 1: Which of the following combinations can result in the formation of a buffer? a) 50 mL of 0.10 M HF with 50 mL of 0.10 M NaOH. b) 50 mL of 0.10 M HNO2 with 25 mL of 0.10 M Ca(OH)2. C) 50 mL of 0.10 M CH3CO2H with 60 mL of 0.10 M NaOH. d) 50 mL of 0.10 M HF with 30 mL of 0.10 M NaOH.

please explain.

Answer 1

a)

Here mol of HF and NaOH is equal.

Both will react and form F-.

But there will not be any HF.

So, it cant form buffer

b)

Here, mol of HNO2 and mol of OH- is same.

So, all HNO2 will react

So, it cant form buffer

c)

Mol of NaOH is more than mol of CH3CO2H.

So, finally, there will be NaOH and CH3CO2Na.

They are both basic.

They will not form buffer

d)

Amount of base is less than amount of HF.

So, some of HF will react with all of NaOH to form some F-.

At the end there would be HF and F-.

It form basic buffer.

Answer: d