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Not to make duplicate, I will refer to my answer in the mentioned Find the ph and amphiprotic salt is added to water at room temperature question.

Of the buffering

If $\mathrm{p}K_\mathrm{a1}$ is very close to $ \mathrm{p}K_\mathrm{a2}$, then the acidic salt solution is a good double buffer $$ \ce{H2A/HA-}, \ce{HA^-/A^2-}$$

But if $\mathrm{p}K_\mathrm{a1} \lt\lt \mathrm{p}K_\mathrm{a2}$, then both buffer systems are far from their $\mathrm{p}K_\mathrm{a}$.

As the consequence, they have just fraction of the buffer capacity, compared to the case with close $\mathrm pK_\mathrm a$ values. We have then a mix of 2 inferior buffers.

If the $\mathrm{p}K_\mathrm{a}$ difference = 4,
the buffer capacity is roughly 25 times smaller.

( 2 orders=100, but 2 buffers ->50, and 0.5/0.5 vs 0.99:0.01 --> 25 )

Not to make duplicate, I will refer to my answer in the mentioned Find the ph and amphiprotic salt is added to water at room temperature question.

Of the buffering

If $\mathrm{p}K_\mathrm{a1}$ is very close to $ \mathrm{p}K_\mathrm{a2}$, then the acidic salt solution is a good double buffer $$ \ce{H2A/HA-}, \ce{HA^-/A^2-}$$

But if $\mathrm{p}K_\mathrm{a1} \lt\lt \mathrm{p}K_\mathrm{a2}$, then both buffer systems are far from their $\mathrm{p}K_\mathrm{a}$.

As the consequence, they have just fraction of the buffer capacity, compared to the case with close $\mathrm pK_\mathrm a$ values. 

We have then a mix of 2 inferior buffers.

If e.g.the $\mathrm{p}K_\mathrm{a}$ difference = 4, the buffer capacity is roughly 25 times smaller, compared to $\mathrm pH=\mathrm pK_\mathrm {a1}$ or $\mathrm pH=\mathrm pK_\mathrm {a2}$.

( 2 orders=100, but 2 buffers ->50, and 0.5/0.5 vs 0.99:0.01 --> 25 )

Not to make duplicate, I will refer to my answer in the mentioned Find the ph and amphiprotic salt is added to water at room temperature question.

Of the buffering

If $\mathrm{p}K_\mathrm{a1}$ is very close to $ \mathrm{p}K_\mathrm{a2}$, then the acidic salt solution is a good double buffer $$ \ce{H2A/HA-}, \ce{HA^-/A^2-}$$

But if $\mathrm{p}K_\mathrm{a1} \lt\lt \mathrm{p}K_\mathrm{a2}$, we have mix of 2 inferior buffers, where each system is far from its $\mathrm{p}K_\mathrm{a}$. As the consequence, they have just fraction of the buffer capacity of the case with close $\mathrm pK_\mathrm a$ values.

If the $\mathrm{p}K_\mathrm{a}$ difference = 4,
the buffer capacity is roughly 25 times smaller.

( 2 orders=100, but 2 buffers ->50, and 0.5/0.5 vs 0.99:0.01 --> 25 )

Not to make duplicate, I will refer to my answer in the mentioned Find the ph and amphiprotic salt is added to water at room temperature question.

Of the buffering

If $\mathrm{p}K_\mathrm{a1}$ is very close to $ \mathrm{p}K_\mathrm{a2}$, then the acidic salt solution is a good double buffer $$ \ce{H2A/HA-}, \ce{HA^-/A^2-}$$

But if $\mathrm{p}K_\mathrm{a1} \lt\lt \mathrm{p}K_\mathrm{a2}$, then both buffer systems are far from their $\mathrm{p}K_\mathrm{a}$. 

As the consequence, they have just fraction of the buffer capacity, compared to the case with close $\mathrm pK_\mathrm a$ values. We have then a mix of 2 inferior buffers.

If the $\mathrm{p}K_\mathrm{a}$ difference = 4,
the buffer capacity is roughly 25 times smaller.

( 2 orders=100, but 2 buffers ->50, and 0.5/0.5 vs 0.99:0.01 --> 25 )

Not to make duplicate, I will refer to my answer in the mentioned Find the ph and amphiprotic salt is added to water at room temperature question.

Of the buffering

If $\mathrm{p}K_\mathrm{a1}$ is very close to $ \mathrm{p}K_\mathrm{a2}$, then the acidic salt solution is a good double buffer $$ \ce{H2A/HA-}, \ce{HA^-/A^2-}$$.

But if $\mathrm{p}K_\mathrm{a1} \lt\lt \mathrm{p}K_\mathrm{a2}$, we have mix of 2 inferior buffers, where each system is far from its $\mathrm{p}K_\mathrm{a}$, having just fraction of the buffer capacity of the former.

If the $\mathrm{p}K_\mathrm{a}$ difference = 4,
the buffer capacity is roughly 25 times smaller.

( 2 orders=100, but 2 buffers ->50, and 0.5/0.5 vs 0.99:0.01 --> 25 )

Not to make duplicate, I will refer to my answer in the mentioned Find the ph and amphiprotic salt is added to water at room temperature question.

Of the buffering

If $\mathrm{p}K_\mathrm{a1}$ is very close to $ \mathrm{p}K_\mathrm{a2}$, then the acidic salt solution is a good double buffer $$ \ce{H2A/HA-}, \ce{HA^-/A^2-}$$

But if $\mathrm{p}K_\mathrm{a1} \lt\lt \mathrm{p}K_\mathrm{a2}$, we have mix of 2 inferior buffers, where each system is far from its $\mathrm{p}K_\mathrm{a}$. As the consequence, they have just fraction of the buffer capacity of the case with close $\mathrm pK_\mathrm a$ values.

If the $\mathrm{p}K_\mathrm{a}$ difference = 4,
the buffer capacity is roughly 25 times smaller.

( 2 orders=100, but 2 buffers ->50, and 0.5/0.5 vs 0.99:0.01 --> 25 )