Question

Which of the following statement(s) is/are correct for a salt of weak acid and weak base in an aqueous solution at ${25}^{\circ }C$.

Here, $K_a$ is the dissociation constant for a weak acid and $K_b$ is the dissociation constant for a weak base.

• A. If $K_a>K_b$ then the solution is neutral.
• B. If $K_a<K_b$ then the solution is acidic in nature.
• C. If $K_a>K_b$ then the solution is acidic in nature.
• D. If $K_a<K_b$ then the solution is alkaline in nature.

Answer 1

Answer: (C), (D)

If $K_a<K_b$ then the solution is alkaline in nature.

For an aqueous solution of weak acid and weak base at ${25}^{\circ }C$,
$pH=7+\frac{1}{2}(p{K}_a-p{K}_b)$
Since , $p{K}_a=-log\left(K_a\right)p{K}_b=-log\left(K_b\right)$
Putting in the $pH$ equation,
$pH=7+\frac{1}{2}(-log(K_a)+log(K_b\left)\right)pH=7+\frac{1}{2}\left(log\right(K_b)-log(K_a\left)\right)$
When, $K_a>K_{b}log\left(K_a\right)>log\left(K_b\right)\Rightarrow pH<7$.
$\therefore$ Solution is acidic in nature.

When,
$K_a<K_{b}log\left(K_a\right)<log\left(K_b\right)\Rightarrow pH>7$
Solution is alkaline in nature.


Theory :
For a weak acid and weak base
$-log\left[H^{+}\right]=-\frac{1}{2}log{K}_w-\frac{1}{2}log{K}_a+\frac{1}{2}log{K}_b$
$pH=\frac{1}{2}(p{K}_w+p{K}_a-p{K}_b)$
valid only if h<0.1 or 10%
$pH=7+\frac{1}{2}(p{K}_a-p{K}_b)$ at ${25}^{o}C$
Since expression for $\left[H^{+}\right]$ or pH is independent of concentration of salt.
So, if $K_a=K_b=pH=7$ it is a neutral solution.
If $K_a>K_b=pH<7$ it is an acidic solution.
If $K_a<K_b=pH>7$ it is a basic solution.

$pH=\frac{1}{2}(p{K}_w+p{K}_a-p{K}_b)$
This formula is always valid for any $K_{a}and{K}_b$ at any temperatures and for any values of h.
pH is independent of the concentration of the salt solution.