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Determine whether or not the mixing of each of the two solutions indicated below will result in a buffer.
A. 75.0 mL of 0.10 MHF; 55.0 mL of 0.15 MNaF
B. 150.0 mL of 0.10 MHF; 135.0 mL of 0.175 MHCl
C. 165.0 mL of 0.10 MHF; 135.0 mL of 0.050 MKOH
D. 125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 MCH3NH3Cl
E. 105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl

Answer :

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Part A

75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF

This combination will form a buffer.

Explanation

Here, weak acid HF and its conjugate base F- is available in the solution

Part B

150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl

This combination cannot form a buffer.

Explanation

Here, moles of HF = 0.15 x 0.1 = 0.015 moles

Moles of HCl = 0.135 x 0.175 = 0.023

Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution

Part C

165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH

This combination will form a buffer.

Explanation

Moles of HF = 0.165 x 0.1 = 0.0165 moles

Moles of KOH = 0.135 x 0.05 = 0.00675 moles

Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer

Part D

125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl

This combination will form a buffer

Explanation

Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer

Part E

105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl

This combination will form a buffer

Explanation

Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles

Moles of HCl = 0.095 x 0.1 = 0.0095 moles

Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer
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A buffer is a solution that mitigates against changes in acidity/alkalinity.

A buffer consists of a weak acid and its conjugate base. Also, a buffer can be formed from a weak base and its conjugate acid. A buffer is a solution that helps to mitigate against changes in acidity and alkalinity.

Let us now examine the solution mixtures listed in the question:

  • 75.0 mL of 0.10 MHF; 55.0 mL of 0.15 MNaF. This can work as a buffer solution because it contains a weak acid (HF) and its conjugate base(F^-).
  • 150.0 mL of 0.10 MHF; 135.0 mL of 0.175 MHCl will not function as a buffer solution
  • 165.0 mL of 0.10 MHF; 135.0 mL of 0.050 MKOH will not function as a buffer solution
  • 125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 MCH3NH3Cl will not function as a buffer solution
  • 105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl will not function as a buffer solution.

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