What amounts of a weak acid and a strong base, when combined, will lead to the formation of a buffer?

Question

BlackTom AI · Accepted Answer

To determine which mixture forms a buffer, we need a solution that contains both a weak acid (HA) and its conjugate base (A-) in appreciable amounts. A buffer typically arises when a weak acid has been partially neutralized by a strong base, leaving both HA and A- present.
Option 1: '1 mol weak acid + 1 mol strong base' — The strong base will fully react with the 1 mole of weak acid: HA + OH- → A- + H2O. This leaves only the conjugate base (A-) in solution and no weak acid, so it cannot act as a buffer. This option fails to provide the necessary HA/A- pair.
Option 2: '1 mol weak acid + 2 mol strong base' — Here, two moles of OH- would neutralize more than the available HA (assuming it's the same HA), driving the reaction past partial neutralization and consuming all HA and creating excess base (A- and OH-). The mixture would not be a typical HA/A- buffer, and the excess base would dominate, so this is not a buffer scenario.
Option 3: '2 mol weak acid + 1 mol strong base' — The strong base neutralizes half of the weak acid: HA + OH- → A- + H2O. After reaction, you have 1 mol HA remaining and 1 mol A- formed, giving a mixture containing both the weak acid and its conjugate base in roughly equal amounts. This is the classic buffer composition for a weak acid–base conjugate pair, providing resistance to pH change.
Option 4: 'You cannot make a buffered solution from a weak acid and a strong base.' — This statement is overly broad and incorrect. While a strong base alone would not yield a buffer with the weak acid, partial neutralization as in option 3 does create a buffer. Buffers do not require only weak base components; they require a weak acid/base conjugate pair in the right proportions. Hence, this option is misleading and incorrect.

What amounts of a weak acid and a strong base, when combined, will lead to the formation of a buffer?单项选择题

类似问题

A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in Na. Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl. The Ka for HF is 3.5 × 10-4.

A 62.9 kg lady is standing on a weighing scale in an elevator as shown in the diagram below. Assume gravitational acceleration = 15 ms-2 Given that the elevator is accelerating upwards at 3.7 ms-2 , determine the magnitude of the normal force on the lady.

An object is fired from ground level at an angle of 31.9 degrees to the horizontal. Taking acceleration due to gravity = 7.65 ms-2 If the initial speed is 5.6 ms-1, how far from the starting point will it hit the ground? Assume air resistance is negligible.

Which of the following equations of motion would be used to calculate the displacement given the initial and final velocities as well as the acceleration?

A 51.8 kg lady is standing on a weighing scale (calibrated to 9.81 ms-2) in an elevator as shown in the diagram below. The scale is then calibrated to 7.67 ms-2 Given that the elevator is accelerating downwards at 2.2 ms-2 , determine the reading on the scale.

An object is fired from ground level at an angle of 39 degrees to the horizontal. Taking acceleration due to gravity = 7.65 ms-2 If the initial speed is 6.8 ms-1, how far from the starting point will it hit the ground? Assume air resistance is negligible.

Which of the following equations of motion would be used to calculate the displacement given the initial and final velocities as well as the time taken?

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单