Question

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 1

Only options A (mix of weak acid HF and its conjugate base NaF) and D (mix of weak base CH₃NH₂ and its conjugate acid CH₃NH₃Cl ) will result in buffer solutions. Options B, C, and E involve either strong acids or bases that would disrupt the buffer system.

Step-by-step explanation:

To determine whether the mixing of each solution will result in a buffer, we need to identify solutions that contain a weak acid along with its conjugate base or a weak base along with its conjugate acid. Buffers are best when the concentrations of the acid/base and its conjugate are comparable.

• Option A: HF is a weak acid and NaF is its conjugate base. Mixing these two solutions will result in a buffer solution.
• Option B: Mixing HF with HCl will not create a buffer because HCl is a strong acid and will not contribute to a buffer system with HF.
• Option C: HF is a weak acid and KOH is a strong base. The resulting mixture will not be a buffer because the strong base will neutralize the weak acid, leaving an excess of hydroxide ions.
• Option D: CH₃NH₂ is a weak base and its conjugate acid is CH₃NH₃Cl. Mixing these will create a buffer solution.
• Option E: Mixing a weak base CH₃NH₂ with a strong acid HCl will not result in a buffer, as the strong acid will likely neutralize the weak base completely.

Based on this analysis, only the mixtures in Options A and D will create buffer solutions.

Answer 2

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