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Question: The dissociation constant (\({{K}_{a}}\)) and percent degree of dissociation (\(\alpha \)) of a weak...

The dissociation constant (Ka{{K}_{a}}) and percent degree of dissociation (α\alpha ) of a weak monobasic acid solution of 0.1M0.1M with a pH=5pH=5 are respectively:
a)10−9,1 b)10−9,10−4 c)10−9,10−2 d)10−5,10−2 \begin{aligned} & a){{10}^{-9}},1 \\\ & b){{10}^{-9}},{{10}^{-4}} \\\ & c){{10}^{-9}},{{10}^{-2}} \\\ & d){{10}^{-5}},{{10}^{-2}} \\\ \end{aligned}

Explanation

Solution

Hint : A monobasic acid is one that can convert into its conjugate base by donating one proton to an external agent. The given pHpH can be used to calculate the concentration of protons in the solution. The percent degree of dissociation is a ratio.

Complete step by step solution :
Since a monobasic acid can only donate one Hydrogen atom to a base, we can think of its general formula to be HXHX.
The given acid is a weak one, which means it will never dissociate completely. The reaction will be in a dynamic equilibrium. It is as follows:
HX⇌H++X−HX\rightleftharpoons {{H}^{+}}+{{X}^{-}}
The concentration of proton and the conjugate base are the same, so let us assume that to be “x”.
Since the pH of the given acid is 5, we can calculate the amount of dissociated Hydrogen ion obtained as follows:
5=−log⁡(x) ⇒x=10−5 \begin{aligned} & 5=-\log \left( x \right) \\\ & \Rightarrow x={{10}^{-5}} \\\ \end{aligned}
Hence the concentration of “H+{{H}^{+}}” and “X−{{X}^{-}}” is 10−5M{{10}^{-5}}M.
We know that the dissociation constant Ka{{K}_{a}} of a weak acid HX is given by,
Ka=[H+][X−][HX]{{K}_{a}}=\dfrac{[{{H}^{+}}][{{X}^{-}}]}{[HX]}
Putting the respective values in the above equation we get:
Ka=(10−5)20.1=10−9{{K}_{a}}=\dfrac{{{\left( {{10}^{-5}} \right)}^{2}}}{0.1}={{10}^{-9}}
So the value of dissociation constant is 10−9{{10}^{-9}}.
Now, dissociation percentage is given by:
Moles of dissociated ionsTotal number of moles×100\dfrac{Moles\text{ of dissociated ions}}{Total\text{ number of moles}}\times 100
Putting the values, the above equation becomes:
10−50.1×100=0.01\dfrac{{{10}^{-5}}}{0.1}\times 100=0.01%
The dissociation percent is therefore 10−2{{10}^{-2}}and the dissociation constant is 10−9{{10}^{-9}}.
Therefore, we can conclude that the answer to this question is option (c)

Note : The compound given here was a weak acid. We always have to represent the dissociation of weak acids in equilibrium. The strong acids such as the mineral acids, dissociate completely and therefore they do not require a representation at equilibrium. The dissociation per cent for a strong acid is always hundred percent.