Such a solution consists of a mixture of Weak Acid and it's salt with a strong base such that the acid & salt concentration are related as

`1.0 ge ([SALT])/([ACID]) ge 0.1`

For eg., Mixture of `CH_3COOH + CH_3COONa`

Mixture of `CH_3COOH + CH_3COOK`

Mixture of `HCN + NaCN`

Mixture of `H_3PO_4+ NaH_2PO_4`

`text(Explanation of Buffer Action -)`

Let us consider a buffer solution made up of `CH_3COOH` and `CH_3COONa`. The weak acid dissociates to a very small extent due to the common ion effect of its salt.

`CH_3COOH ⇋ CH_3COO^(-) + H^(+)`

`CH_3COONa ⇋ CH3COO^(-) + Na^+`

Now let us assume that this solution contains `10` moles of `CH_3COOH` and `10` moles of `CH_3COONa`. The salt being a strong electrolyte would completely dissociate while the acid would be hardly dissociated. We assume that the amount of `CH_3COO^(-)` is `10` moles as the contribution from `CH_3COOH` would be negligible. We also assume that the amount of `CH_3COOH` to be `10` moles as it would be very weakly dissociated. Therefore, the solution contains `10` moles of `CH_3COOH` and `10` moles of `CH3COO^-` ions.

Let us now add `1` mole of `H^+` ions to this solution. These `1` mole of `H^+` ions would react with the `1` mole of `CH_3COO^-` ions to produce `1` mole of `CH_3COOH`. (This is because the reaction of `CH_3COOH` to give `CH_3COO^-` ions and `H^+` ions has an equilibrium constant value of approximately `10^(-5)`. Therefore, the reverse reaction, that is the reaction of `CH_3COO^(-)` ions to combine with `H^+` ions to give `CH_3COOH` would have an equilibrium constant of approximately `10^5`. So, this reaction can be assumed to be complete). This `1` mole of `CH_3COOH` formed would ionize weakly because it is a weak acid & also due to the common ion effect of `CH_3COO^-` ions. So, the amount of `H^+` ions produced back is much less than the `1` mole that was added. So, effectively, `1` mole of `H^+` ions was consumed and an amount much less than that is produced back which causes an insignificant change in `pH` of the buffer solution.

Similarly, let us add `1` mole of `NaOH` ions to the same buffer solution containing `10` moles of `CH_3COOH` and `10` moles of `CH_3COO^-` ions. These `1` mole of `OH^-` ions added would be consumed by the `1` mole of `CH_3COOH` to produce `1` mole of `CH_3COO^-` ions. The `CH_3COO^-` ions produced would be hydrolyzed weakly and thus the `OH^-` ion produced is much less than the amount that was added. This causes a minor change in `pH`.

Such a solution consists of a mixture of Weak base and it's salt with a strong acid with the base & salt concentration related as

`1.0 ge ([SALT])/([BASE]) ge 0.1`

For eg., Mixture of `NH_4OH + NH_4Cl`

Mixture of `C_5H_5N` (Pyridine) `+ C_5H_6N^(+)Cl^(-)` (Pyridinium Chloride)


`text(Explanation of Buffer Action -)`

It is similar to that of an Acidic Buffer, the only difference is that in Basic Buffer the addition of Acid(`H^+`) is countered by the Weak Base and the addition of a Base(`OH^-`) is countered by the salt.

`text(pH of an Basic Buffer -)`

e.g., `NH_4OH + NH_4Cl` Buffer

Similar to an Acidic Buffer,

For Basic Buffers, `pOH = pK_b + log([[text(Salt)]]/[[text(Base)]])`

A salt buffer is a solution of a salt which itself can act as a buffer. Such salt is the salt of a weak acid and weak base.

For example in an aqueous solution of `CH_3COONH_4`

`CH_3COONH_4 ⇋ CH_3COO^(-) + NH_4^(+)`

When acid is added to this solution, `CH_3COO^-` ion consumes it to give `CH_3COOH` (Weak Acid) & when a
base is added, `NH_4^(+)` ion consumes it to give `NH_4OH` (Weak Base). Hence it acts as a buffer.

For such solutions, `[H^(+)]= sqrt ((K_w K_a)/K_b)`

`pH=1/2 pK_w + 1/2 pK_a -1/2 pK_b`

Buffer capacity is defined as the amount (moles) of a strong acid or strong base required to change the `pH` of a buffer solution by one unit. Let there be a buffer solution of volume `1` L with `b` moles of salt and `a` moles of acid. The `pH` of the buffer would be given by

`pH = pK_a + log (b/a)`

On adding, `x` moles of a strong acid (monobasic), the `pH` changes to

`pH' = pK_a + log ((b-x)/(a+x))`

`pH' = pH - log (b/a) + log ((b-x)/(a+x))`

`Delta(pH) = log [(b/a)xx((a+x)/(b-x))] = 1/2.303 ln[(b/a)xx((a+x)/(b-x))]`

Differentiating with respect to `x` we get

`(dDelta(pH))/dx = (1/[(b/a)xx((a+x)/(b-x))]) xx (b/a) xx([(b-x)xx1 + (a+x)]/(b-x)^2) xx(1/2.303)`

Taking the inverse

`dx/(dDelta(pH)) = 2.303 xx(((a+x)(b-x))/(a+b))`

This is defined as buffer capacity It is the ratio of the small amount of acid or base added to the change in `pH` caused in the buffer.

It can be proved that the maximum buffer capacity is achieved when the salt and acid or base concentration is equal. Differentiating buffer capacity with respect to `b`, the amount of salt present in the buffer and equating it to zero, gives

`d/(db)(dx/(dDelta(pH)) ) =0`

On Solving, we get `b - a + 2 x = 0`; since `x` is very small we ignore `2x`
Thus, `b - a = 0 => b = a`

The buffer shows maximum buffer capacity when the amount of acid or base and the salt are same.

It is difficult to give an exact limit upto which a buffer can be used, it in generally accepted that a solution has useful buffer capacity provided that the value of [Salt]/[Acid] lie within the range of `0.1` to `10`. Therefore, the `pH` range of a buffer solution is generally specified
as `pK_a � 1` (for acidic buffers) & `pK_b � 1` (for basic buffers)