Question

Correct relation b/w dissociation constant's of a di-basic acid:

• A. \(Ka_1=Ka_2\)
• B. \(Ka_1>Ka_2\)
• C. \(Ka_1<Ka_2\)
• D. \(Ka_1=\frac {1}{Ka_2}\)

Answer 1

The Correct Option is B

 To solve this question, we need to understand the concept of dissociation constants for a di-basic acid.

A di-basic acid is an acid with two acidic hydrogens to donate sequentially. The dissociation process happens in two steps, each with its own dissociation constant:

\[HA_2 \rightleftharpoons H^+ + HA^-\]

with the dissociation constant \(Ka_1\)

\[HA^- \rightleftharpoons H^+ + A^{2-}\]

with the dissociation constant \(Ka_2\)

In most di-basic acids, the first dissociation is stronger than the second. This is because once the first hydrogen is lost, the negative charge on the anion \(HA^-\) attracts the hydrogen ion more strongly, making it harder for the acid to lose the second hydrogen ion. Thus, \(Ka_1\) is generally greater than \(Ka_2\).

Therefore, the correct relation between the dissociation constants of a di-basic acid is:

\(Ka_1 > Ka_2\)

This makes option 2, \(Ka_1 > Ka_2\), the correct answer.

Let's also review why the other options are incorrect:

• \(Ka_1 = Ka_2\): This implies equal dissociation ability for both steps, which is generally not the case in di-basic acids.
• \(Ka_1 < Ka_2\): This suggests the second dissociation is stronger, which is even more unlikely considering the increased electrostatic attraction once the first hydrogen is lost.
• \(Ka_1 = \frac{1}{Ka_2}\): This doesn't hold any standard relationship known for di-basic acids.