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Question: Calculate the ratio of \(HCO{{O}^{-}}\) and \({{F}^{-}}\) in a mixture of 0.2 M HCOOH (\({{K}_{a}}=2...

Calculate the ratio of HCOO−HCO{{O}^{-}} and F−{{F}^{-}} in a mixture of 0.2 M HCOOH (Ka=2 x 10−4{{K}_{a}}=2\text{ x 1}{{\text{0}}^{-4}}) and 0.1 M HF (Ka=6.6 x 10−4{{K}_{a}}=6.6\text{ x 1}{{\text{0}}^{-4}})
(a). 1: 6.6
(b). 1: 3.3
(c). 2: 3.3
(d). 3.3: 2

Explanation

Solution

First we have to write the equation for both the compound when dissolved in water, then we write the equation of equilibrium constant. Now, we can compare both the value of the equilibrium constant, we can find the ratio of the ions.

Complete answer:
We are given two compounds, HCOOH and HF, let write the dissociation equation for both the compounds when dissolved in water.
HCOOH+H2O⇌HCOO−+H3O+HCOOH+{{H}_{2}}O\rightleftharpoons HCO{{O}^{-}}+{{H}_{3}}{{O}^{+}}
Let us assume that before equilibrium, the concentration of HCOOH is C and after equilibrium, the concentration of HCOOH, HCOO−HCO{{O}^{-}} and H3O+{{H}_{3}}{{O}^{+}} are C-x, x, and x respectively.
HF+H2O⇌F−+H3O+HF+{{H}_{2}}O\rightleftharpoons {{F}^{-}}+{{H}_{3}}{{O}^{+}}
Let us assume that before equilibrium, the concentration of HF is D and after equilibrium, the concentration of HF, F−{{F}^{-}} and H3O+{{H}_{3}}{{O}^{+}} are D-y, y, and y respectively.
Due to the common ion effect, the concentration of H3O+{{H}_{3}}{{O}^{+}} will be x+y in both the equations. Since the value is acid constants for both compounds are very less, hence we can neglect x in case of C-x for HCOOH and y in case of D-y for HF
Now, writing the equilibrium constant for both the reactions.
Ka(HCOOH)=x(x+y)C{{K}_{a}}(HCOOH)=\dfrac{x(x+y)}{C}
Ka(HF)=y(x+y)D{{K}_{a}}(HF)=\dfrac{y(x+y)}{D}
When we divide these equations, we get:
xy=Ka(HCOOH)CKa(HF)D\dfrac{x}{y}=\dfrac{{{K}_{a}}(HCOOH)C}{{{K}_{a}}(HF)D}
As we know that x and y are HCOO−HCO{{O}^{-}}and F−{{F}^{-}} respectively, and the value of acid constants and the value of concentration of both the ions are given, we can put the values:
xy=[HCOO−][F−]=2 x 10−4 x 0.26.6 x 10−4 x 0.1\dfrac{x}{y}=\dfrac{[HCO{{O}^{-}}]}{[{{F}^{-}}]}=\dfrac{2\text{ x 1}{{\text{0}}^{-4}}\text{ x 0}\text{.2}}{6.6\text{ x 1}{{\text{0}}^{-4}}\text{ x 0}\text{.1}}
[HCOO−][F−]=23.3\dfrac{[HCO{{O}^{-}}]}{[{{F}^{-}}]}=\dfrac{2}{3.3}
Therefore, the ratio is 2: 3.3

So, the correct answer is “Option c”.

Note:
The formula of the equilibrium constant is written as the ratio of the product of the concentration of products by the product of the concentration of the reactants. You can only neglect the value of concentration when its value is very less.