Question

Buffer capacity of acid buffer solution is more when
A. \[pK_a = pH\]
B. $[salt] = [acid]$
C. \[pK_a = 7\]
D. \[[H + ] = pK_a\]

1. All are correct.
2. C and D are correct.
3. A and B are correct.
4. C and B are correct.

Answer 1

Hint: To solve this problem we should have knowledge of acidic and basic buffer. There are various types of buffers and there exist a relation between the \[pH\] and \[pOH\] of different type of buffers with $p{K_a}$, $p{K_b}$, salt, acid and base concentration.

Formula Used:
$pH = p{K_a} + \log \dfrac{{[Salt]}}{{[Acid]}}$

Complete answer:
Let us understand the buffer solution in detail to know the concept behind the buffers.
Buffer solutions are those, which resist a change in \[pH\] upon addition of a small amount of acid or base. This does not mean that the \[pH\] will not change, and all it means is that the \[pH\] change would be less than the change that would have occurred had it not been a buffer.
There are various types of buffers:
Buffer of weak acid and its salt with a strong base.
Buffer of a weak base and its salt with a strong acid.
The solution of the salt of weak acid and a weak base.
To calculate the \[pH\]of a buffer solution made up of a weak acid and its salt with a strong base. We have
$C{H_3}COOH \rightleftarrows C{H_3}CO{O^ - } + {H^ + }$
$\Rightarrow {K_a}$=$\dfrac{{[CH3CO{O^ - }][{H^ + }]}}{{[CH3COOH]}}$=$\dfrac{{[Salt][{H^ + }]}}{{[Acid]}}$
$ \Rightarrow $$[{H^ + }]$=${K_a} \times \dfrac{{[Acid]}}{{[Salt]}}$
Taking log of both sides.
$\Rightarrow \log [{H^ + }] = \log {K_a} + \log \dfrac{{[Acid]}}{{[Salt]}}$
And we know $pH = - \log ({H^ + })$ and $p{K_a} = - \log ({K_a})$
So we get, $pH = p{K_a} + \log \dfrac{{[Salt]}}{{[Acid]}}$
This is known as the Henderson’s equation of a buffer.
For a buffer made up of weak base and its salt with a strong acid the Henderson’s equation looks like this:
$pOH = p{K_b} + \log \dfrac{{[Salt]}}{{[Base]}}$

Now let us know about buffer capacity.

Buffer capacity: It is defined as the mole of a strong acid or strong base required to change the \[pH\]of a buffer by one unit, for 1L Buffer solution.
Maximum buffer capacity: It can be proved that the maximum buffer capacity is achieved when the salt and acid or base concentration is equal.
So for more buffer capacity 

$\
  [Acid] = [Salt] \\
  pH = p{K_a} \\
\ $

Hence, the correct option is C.

Note: Remember buffer capacity is optimal when the ratio is 1:1; that is, when $pH = p{K_a}$ and $p{K_a}$ will not affect the buffer capacity. Learn Henderson's equations very carefully to solve similar types of questions.