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."]
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]
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