Question

Explain why a solution that is 1.3 M HF and 1.3 mM KF is not a good buffer. HF is a strong acid and cannot be used in a buffer system. The ratio of acid to conjugate base is outside the buffer range of 10:1. The two species are not a conjugate acid base pair. KF is not soluble in water and cannot be used in a buffer system.

Answer 1

Answer:

The ratio of acid to conjugate base is outside the buffer range of 10:1.

Explanation:

The Henderson-Hasselbalch equation for a buffer is

$\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{\text{[A ^{-} ]}}{\text{[HA]}}$

A buffer should have

$\dfrac{1}{10} \leq \dfrac{\text{[A ^{-}] }}{\text{[HA]}} \leq \dfrac{10}{1}$

For a solution that is 1.3 mol·L⁻¹ in HF and 1.3 mmol·L⁻¹ in KF, the ratio is

$\dfrac{1.3 \times 10^{-3} }{1.3} = \dfrac{1}{1000}$

The ratio of acid to conjugate base is 1000:1, which is outside the range of 10:1.

A is wrong. NF is a weak acid.

C is wrong. The two species are a conjugate acid-base pair.

D is wrong. Salts of Group 1 metals are soluble.

Answer 2

The solution is not a good buffer as the ratio of acid to the conjugate base is outside the buffer range of 10:1. Thus statement B is correct.

The ratio of acid to the conjugate base can be calculated as:

pH = $\rm pK_a$ + log $\rm \dfrac{A^-}{HA}$

For a solution that is 1.3 $\rm mol.L^-^1$ in HF and 1.3 $\rm mmol.L^-^1$ in KF, the ratio is:

$\rm \dfrac{1.3\;\times\;10^-^3}{1.3}\;=\;\dfrac{1}{1000}$

The ratio of acid to the conjugate base has been equal to 1:1000.

The ratio of acid to the conjugate base is outside the buffer range of 10:1. Thus the solution is not a good buffer. Thus statement B is correct.

For more information about the buffer system, refer to the link:

https://brainly.com/question/1423164