Question

What is \[[{{B}^{-}}]\] in a solution that has 0.03 M HA and 0.1 M HB?
\[{{K}_{a}}\]for HA and HB are \[1.38\times {{10}^{-4}}\]and \[1.05\times {{10}^{-10}}\], respectively.
(A) \[5.15\times {{10}^{-9}}M\]
(B) \[5.15\times {{10}^{-3}}M\]
(C) \[5.15\times {{10}^{-5}}M\]
(D) \[5.15\times {{10}^{-7}}M\]

Answer 1

Hint: Acid dissociation constant, \[{{K}_{a}}\]: It is also known as acidity constant and it is a quantitative measure of the strength of an acid present in solution. It is the equilibrium constant for a chemical reaction.
For \[HB\overset{{}}{\longleftrightarrow}{{H}^{+}}+{{B}^{-}}\]
\[{{K}_{a}}=\dfrac{[{{H}^{+}}][{{B}^{-}}]}{[HB]}\]

Complete step by step answer:
The given concentrations of HA and HB in question are 0.03 M and 0.1 M respectively. .
In the question it is given that the values of Acid dissociation constants (\[{{K}_{a}}\]) HA =\[1.38\times {{10}^{-4}}\]and \[{{K}_{a}}\]of HB =\[1.05\times {{10}^{-10}}\]
HA undergoes dissociation as follows.
For \[HA\overset{{}}{\longleftrightarrow}{{H}^{+}}+{{A}^{-}}\]
Formula to calculate acid dissociation constant (\[{{K}_{a}}\]) for the above equation is as follows.
\[{{K}_{a}}=\dfrac{[{{H}^{+}}][{{A}^{-}}]}{[HA]}\]
Now substitute all the known values in the above equation to get the concentration of \[[{{H}^{+}}]\].
\[\begin{align}
  & {{K}_{a}}=\dfrac{[{{H}^{+}}][{{A}^{-}}]}{[HA]} \\
 & 1.38\times {{10}^{-4}}=\text{ }\dfrac{{{[{{H}^{+}}]}^{2}}}{[HA]} \\
 & {{[{{H}^{+}}]}^{2}}=\text{ }1.38\times {{10}^{-4}}\times 0.03 \\
 & [{{H}^{+}}]=0.002M \\
\end{align}\]
From the above calculation the concentration of the\[[{{H}^{+}}]\]is 0.002M.
HB undergoes dissociation as follows
For \[HB\overset{{}}{\longleftrightarrow}{{H}^{+}}+{{B}^{-}}\]
Formula to calculate acid dissociation constant for the above equation is as follows.
\[{{K}_{a}}=\dfrac{[{{H}^{+}}][{{B}^{-}}]}{[HB]}\]
Now substitute all the known values in the above equation to get the concentration of \[[{{B}^{-}}]\].
\[\begin{align}
  & {{K}_{a}}=\dfrac{[{{H}^{+}}][{{B}^{-}}]}{[HB]} \\
 & 1.05\times {{10}^{-10}}=\text{ }\dfrac{[{{H}^{+}}][{{B}^{-}}]}{[HA]} \\
 & [{{B}^{-}}]=\text{ }\dfrac{1.05\times {{10}^{-10}}\times 0.1}{0.002} \\
 & [{{B}^{-}}]=5.15\times {{10}^{-9}}M \\
\end{align}\]
From the above calculation the concentration of the \[[{{B}^{-}}]\]is \[5.15\times {{10}^{-9}}M\].
Coming to option A, \[5.15\times {{10}^{-9}}M\]. It is matching with the answer we got.

So, the correct option is A.

Additional information:
\[{{K}_{a}}\]is simply a number that is measured for different acids at different temperatures.
There is a wide difference in \[{{K}_{a}}\]values.
For strong acids, \[{{K}_{a}}\]values are high (greater than 10) and for weaker acids, \[{{K}_{a}}\] are much less.
For strong bases, \[{{K}_{b}}\]values are high (greater than 10) and for weaker acids, \[{{K}_{b}}\] are much less

Note: Don’t be confused in between \[{{K}_{a}}\] and\[{{K}_{b}}\].
\[{{K}_{a}}\] is called as dissociation constant of an acid and
\[{{K}_{b}}\] is called the dissociation constant of a base.
The product of the acid dissociation constant and the base-dissociation constant is the ion-product constant for water \[{{K}_{w}}={{K}_{a}}\times {{K}_{b}}\]