Question

One litre of 1M solution of an acid HA $({K_a} = {10^{ - 4}} at 25^\circ C)$ has pH = 2. It is diluted by water so that new pH becomes double. The solution was diluted to $5 \times {10^a}$ mL. The value of a is

Answer 1

Hint: The weak acids or weak bases are those which completely do not dissociate into their constituent ions. These acids and bases are considered weak electrolytes. For the dissociation of weak electrolytes like acetic acid, the dissociation constant K is interconnected with the degree of dissociation $\alpha$ by a formula. By using the formula, the value of a can be determined.

Complete step by step answer:
Given,
The molarity of solution is 1M.
The volume is 1 L.
${K_a} = {10^{ - 4}}$
pH = 2
When the pH is 2, the concentration of hydrogen ion ${H^ + }$ is 0.01M.
When the solution is diluted the pH becomes double.
The new pH is 4, the concentration of hydrogen ion ${H^ + }$ is 0.0001M.
The degree of dissociation $\alpha $ is 0.0001
The dissociation constant of weak electrolytes at concentration c can be calculated by using the formula as shown below.
${K_c} = \dfrac{{C{\alpha ^2}}}{{1 - \alpha }}$
This formula is known as Ostwald’s dilution law.
Substitute the value of $\alpha c$ , in the above equation.
${10^{ - 4}} = 0.0001\dfrac{\alpha }{{1 - \alpha }}$
$\alpha = 0.5$
$c = 2 \times {10^{ - 4}}$
To calculate the volume the formula used is shown below.
${c_1}{v_1} = {c_2}{v_2}$
Substitute the values in the above equation.
$1 \times 1 = 2 \times {10^{ - 4}} \times {V_2}$
$\Rightarrow {V_2} = 5000L$
Convert liter into milliliter.
\[5000L = 5 \times {10^6}mL\]

Thus, the value of a is 6.

Note:
The degree of dissociation is dependent on the concentration of electrolyte. As the concentration of electrolytes increases, the value of the degree of dissociation decreases. Ostwald’s dilution law is only applicable to the weak electrolytes. It does not favor strong electrolytes.