The dissociation constant of a weak acid is $1 \times 10^{-4}$ in order to prepare a buffer solution with a $pH = 5$ the [salt] / [Acid] ratio should be :

Question

In an acidic buffer, if the ratio of base to acid concentration is increased by 100 times, then find the increase in pH of the buffer solution.

Answer 1

To solve this question, we need to understand the concept of buffer solutions and how the pH of such a solution is calculated using the Henderson-Hasselbalch equation. A buffer solution is typically made from a weak acid and its salt (conjugate base). The equation to find the pH of a buffer solution is given by:

pH = pK_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right)

Where:

pK_a = -\log(K_a)
K_a is the dissociation constant of the weak acid.

Given in the question, K_a = 1 \times 10^{-4}. We first calculate pK_a:

pK_a = -\log(1 \times 10^{-4}) = 4

We want the buffer solution to have a pH = 5. Using the Henderson-Hasselbalch equation, we plug in the known values:

5 = 4 + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right)

To isolate the logarithmic part, subtract 4 from both sides:

1 = \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right)

Converting the logarithmic equation to its exponential form gives:

10^1 = \frac{[\text{Salt}]}{[\text{Acid}]}

Which simplifies to:

\frac{[\text{Salt}]}{[\text{Acid}]} = 10

This indicates that the ratio of the concentration of the salt to the concentration of the acid should be 10:1.

Thus, the correct answer is 10 : 1.

Answer 2