Question

A buffer can be prepared by mixing two solutions. Determine if each of the following mixtures will result in a buffer solution or not.  

1) Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.05 M KCN [ Select ] ["No, it will not result in a buffer solution.", "Yes, it will result in a buffer solution."]

2) Mixing 100.0 mL of 0.1 M NH₃ with 100.0 mL of 0.05 M (NH₄)₂SO₄ [ Select ] ["Yes, it will result in a buffer solution.", "No, it will not result in a buffer solution."]

3) Mixing 100.0 mL of 0.1 M HNO₃ with 100.0 mL of 0.1 M NaNO₃ [ Select ] ["No, it will not result in a buffer solution.", "Yes, it will result in a buffer solution."]

4) Mixing 100.0 mL of 0.1 M HF with 100.0 mL of 0.05 M NaOH [ Select ] ["Yes, it will result in a buffer solution.", "No, it will not result in a buffer solution."]

5) Mixing 100.0 mL of 0.1 M HF with 100.0 mL of 0.1 M NaOH [ Select ] ["No, it will not result in a buffer solution.", "Yes, it will result in a buffer solution."]

Answer 1

Buffer solution is a defined as , the solution whose PH value doesnot change by adding a little amount of either strong acid or strong base.

Buffer solution is of two types

a) acidic buffer is a mixture of weak acid and its salt with strong base (OR) mixture of weak acid and its conjugate base,

b) basic buffer is a mixture of weak base and its salt with strong acid (OR) mixture of weak base and its conjugate acid.

1) yes it is a buffer solution. [ It is acidic buffer solution]

mixture of 100.0mL of 0.1M HCN and 100.0 mL of 0.05M KCN

It is a mixture of weak acid (HCN) and its salt with strong base (KOH)

HCN + KCN is buffer solution. [ It is a basic buffer soltuion]

2) Yes ,it is a buffer solution

It is a weak base NH3 and its conjugate acid NH4+.

3) No, it is not a buffer because HNO3 is a strong acid.

So it does not forms a buffer.

4) yes. it is buffer solution.

HF = 100 mL of 0.1M

number of moles of HF = 0.1M x 0.100L = 0.01 moles

NaOH = 100.0 mL of 0.05M

number of moles of NaOH = 0.05M x 0.100L = 0.005 moles

number of moles of weak acid is more than the base.

So it forms a buffer solution.

                  HF + NaOH ------------- NaF + H2O

PH = PKa +og[NaF]/[HF]

5) Yes it is buffer solution

Number of moles of HF = 0.1M x 0.100L = 0.01 moles

number of moles of NaOH = 0.1M x 0.100L = 0.01 moles

So it is equivalent point.

at equivalent point

PH= 7+1/2[PKa + logC]