Question

$20mL$ of 0.1 M Weak acid ${{\text{K}}_{a}}={{10}^{-5}}$ is mixed with solution of $10mL$ of 0.3M HCl and $10mL$ of 0.1M NaOH. Find the value of\[\dfrac{[{{A}^{-}}]}{([HA]+[{{A}^{-}}])}\] in the resulting solution.
a.) $-2\times {{10}^{-4}}$
b.) $-2\times {{10}^{-5}}$
c.) $-2\times {{10}^{-3}}$
d.) -0.05

Answer 1

Hint: ${{K}_{a}}$ is dissociation constant for weak acid.
$Ka=\dfrac{[{{H}^{+}}][{{A}^{-}}]}{[HA]}$
Concentration of [${{H}^{+}}$] can be calculated from formula
$\dfrac{M1V1+M2V2}{V1+V2}$
\[\], Where M is Molarity and V is volume. It can be calculated using Molarity and volume of HCl and NaOH.

Complete step by step answer:
Concentration of $[{{H}^{+}}]$ can be calculated from HCl and NaOH as they dissociate completely, and weak acid does not associate completely.
Concentration of $[{{H}^{+}}]$ can be calculated from formula \[\dfrac{{{M}_{1}}{{V}_{1}}+{{M}_{2}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}}\]

\[=\dfrac{10\times 0.3-10\times 0.1}{20+10+10}\]
 \[=0.05\]
Concentration of weak acid will change as volume is diluted $40mL$.
$$ [HA]=\dfrac{20\times 0.1}{40}=\dfrac{2}{40}=0.05$$
As we know, weak acid HA dissociates as
$HA\text{ }\leftrightarrows{{\text{H}}^{+}}\text{ + }{{\text{A}}^{-}}$
Let us assume concentration of $[{{A}^{-}}]$ as $x$ then concentration of $[{{H}^{+}}]$ will be 0.05+ $x$ concentration of HA will be 0.05- $x$.
Due to the common ion effect, $x$ can be neglected with respect to 0.05.
Common ion effect is a phenomenon which is due to presence of common ion, ionization of molecules reduces.

As we know
${{\text{K}}_{a}}$ is dissociation constant of weak acid and is ratio of molar concentration of product to molar concentration of reactant.
\[\] \[Ka=\dfrac{(0.05+x)(x)}{(0.05-x)}\]

As $x$ can be neglected with respect 0.05
       \[Ka=\dfrac{0.05}{0.05}\times x\]
      As given in data, \[{{\text{K}}_{a}}={{10}^{-5}}=x\]
So the concentration of $[{{A}^{-}}]$ is \[{{10}^{-5}}\].

So, now by substituting the values,
\[\dfrac{\text{ }\!\![\!\!\text{ }{{\text{A}}^{\text{- }\!\!]\!\!\text{ }}}}{([HA]+[{{A }^{-}}])}\]\[=\dfrac{{{10}^{-5}}}{(0.05+x)}\]
As $x$ can be neglected with respect to 0.05,
 =\[\dfrac{{{10}^{-5}}}{0.05}\] =\[2\times {{10}^{-4}}\]
 So, Value of $\dfrac{[{{A}^{-}}]}{([HA]+[{{A}^{-}}])}$ for resulting solution is $2\times {{10}^{-4}}$.
So, the correct answer is “Option A”.

Note: As we know concentration of $[{{A}^{-}}]$ can be neglected with respect to concentration of HA ,
the equation can be written as $\dfrac{[{{A}^{-}}]}{[HA]}$, Concentration of $[{{H}^{+}}]=[{{A}^{-}}]=\sqrt{KaC}=\alpha C$
$[{{H}^{+}}]=[{{A}^{-}}]=\sqrt{KaC}=\alpha C$
Where $\alpha $ is the degree of dissociation and C is Concentration.