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

${{K}_{a}}$ is dissociation constant for weak acid.
$Ka=\dfrac{[{{H}^{+}}][{{A}^{-}}]}{[HA]}$
Concentration of [${{H}^{+}}$] can be calculated from formula
$\dfrac{M1V1+M2V2}{V1+V2}$

**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.