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Question: What is \[[{{B}^{-}}]\] in a solution that has 0.03 M HA and 0.1 M HB? \[{{K}_{a}}\]for HA and HB ...

What is [B−][{{B}^{-}}] in a solution that has 0.03 M HA and 0.1 M HB?
Ka{{K}_{a}}for HA and HB are 1.38×10−41.38\times {{10}^{-4}}and 1.05×10−101.05\times {{10}^{-10}}, respectively.
(A) 5.15×10−9M5.15\times {{10}^{-9}}M
(B) 5.15×10−3M5.15\times {{10}^{-3}}M
(C) 5.15×10−5M5.15\times {{10}^{-5}}M
(D) 5.15×10−7M5.15\times {{10}^{-7}}M

Explanation

Solution

Acid dissociation constant, Ka{{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⟷H++B−HB\overset{{}}{\longleftrightarrow}{{H}^{+}}+{{B}^{-}}
Ka=[H+][B−][HB]{{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 (Ka{{K}_{a}}) HA =1.38×10−41.38\times {{10}^{-4}}and Ka{{K}_{a}}of HB =1.05×10−101.05\times {{10}^{-10}}
HA undergoes dissociation as follows.
For HA⟷H++A−HA\overset{{}}{\longleftrightarrow}{{H}^{+}}+{{A}^{-}}
Formula to calculate acid dissociation constant (Ka{{K}_{a}}) for the above equation is as follows.
Ka=[H+][A−][HA]{{K}_{a}}=\dfrac{[{{H}^{+}}][{{A}^{-}}]}{[HA]}
Now substitute all the known values in the above equation to get the concentration of [H+][{{H}^{+}}].

& {{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{aligned}$$ From the above calculation the concentration of the$$[{{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{aligned} & {{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{aligned}$$ From the above calculation the concentration of the $$[{{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}}$$