Solveeit Logo

Question

Question: Buffer capacity of acid buffer solution is more when A. \[pK_a = pH\] B. \([salt] = [acid]\) C...

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

Explanation

Solution

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 pHpH and pOHpOH of different type of buffers with pKap{K_a}, pKbp{K_b}, salt, acid and base concentration.

Formula Used:
pH=pKa+log[Salt][Acid]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 pHpH upon addition of a small amount of acid or base. This does not mean that the pHpH will not change, and all it means is that the pHpH 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 pHpHof a buffer solution made up of a weak acid and its salt with a strong base. We have
CH3COOHCH3COO+H+C{H_3}COOH \rightleftarrows C{H_3}CO{O^ - } + {H^ + }
Ka\Rightarrow {K_a}=[CH3COO][H+][CH3COOH]\dfrac{{[CH3CO{O^ - }][{H^ + }]}}{{[CH3COOH]}}=[Salt][H+][Acid]\dfrac{{[Salt][{H^ + }]}}{{[Acid]}}
\Rightarrow [H+][{H^ + }]=Ka×[Acid][Salt]{K_a} \times \dfrac{{[Acid]}}{{[Salt]}}
Taking log of both sides.
log[H+]=logKa+log[Acid][Salt]\Rightarrow \log [{H^ + }] = \log {K_a} + \log \dfrac{{[Acid]}}{{[Salt]}}
And we know pH=log(H+)pH = - \log ({H^ + }) and pKa=log(Ka)p{K_a} = - \log ({K_a})
So we get, pH=pKa+log[Salt][Acid]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=pKb+log[Salt][Base]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 pHpHof 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=pKa  \ [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=pKapH = p{K_a} and pKap{K_a} will not affect the buffer capacity. Learn Henderson's equations very carefully to solve similar types of questions.