Question

What is $\left[ {{H^ + }} \right]$ in mol/L of a solution that is 0.20M in $C{H_3}COONa$ and 0.10M in $C{H_3}COOH$? ${K_a}$ for $C{H_3}COOH$=$1.8 \times {10^{ - 5}}$
A. $3.5 \times {10^{ - 4}}$
B. $1.1 \times {10^{ - 5}}$
C. $1.8 \times {10^{ - 5}}$
D. $9.0 \times {10^{ - 6}}$

Answer 1

Hint: Le Chatelier’s principle states that if a chemical system at equilibrium is disturbed the system will react in the direction that counteracts the disturbance the stress.
Example:
${\text{C}}{{\text{O}}_{\left( {\text{g}} \right)}}{\text{ + }}{{\text{H}}_{\text{2}}}{{\text{O}}_{\left( {\text{l}} \right)}} \rightleftarrows {\text{C}}{{\text{O}}_{\text{2}}}_{\left( {\text{g}} \right)}{\text{ + }}{{\text{H}}_{{\text{2}}\left( {\text{g}} \right)}}$
If the concentration of the reactant is increased the equilibrium of the reaction shifted to the right side.
If the concentration of the reactant is decreased the equilibrium of the reaction shifted to the left side.
We can calculate the concentration of hydronium ion $\left[ {{H^ + }} \right]$ by using the value of ${K_a}$ and concentrations of $C{H_3}CO{O^ - }$ and concentration of $C{H_3}COOH$.

Formula used: We can calculate the concentration of hydronium ion using the equation,
$\left[ {{H^ + }} \right] = \dfrac{{{K_a}}}{{\left[ {C{H_3}CO{O^ - }} \right]}} \times \left[ {C{H_3}COOH} \right]$
Here,
$\left[ {{H^ + }} \right]$ is the concentration of hydronium ion
${K_a}$ is the value of dissociation constant
$\left[ {C{H_3}CO{O^ - }} \right]$ is the concentration of $C{H_3}CO{O^ - }$
$\left[ {C{H_3}COOH} \right]$ is the concentration of $C{H_3}COOH$

Complete step by step answer:
Given data contains,
The concentration of $C{H_3}COOH$ is $0.10\,M$.
The concentration of $C{H_3}COONa$ is $0.20\,M$.
The value of ${K_a}$ for $C{H_3}COOH$is $1.8 \times {10^{ - 5}}$.
We can write the ionization equation for $C{H_3}COOH$ as,

We can write the ionization equation for $C{H_3}COONa$ as,
$C{H_3}COONa\xrightarrow{{}}C{H_3}CO{O^ - } + N{a^ + }$
We can that $C{H_3}CO{O^ - }$ ions are common in both the equations, and the equilibrium will be shifted towards the left according to Le Chateliers’s principle.
We know that equilibrium constant is rate of the products divided by the rate of the reactants.
$
  {{\text{K}}_{\text{a}}}{\text{ = }}\dfrac{{\left[ {{\text{rate}}\,{\text{of}}\,{\text{products}}} \right]}}{{\left[ {{\text{rate}}\,{\text{of}}\,{\text{reactants}}} \right]}} \\
  {K_a} = \dfrac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}} \\
 $
Let us now substitute the values of $C{H_3}CO{O^ - }$, $C{H_3}COOH$ and ${K_a}$ to calculate the value of concentrations of $\left[ {{H^ + }} \right]$ ions.
$
  {K_a} = \dfrac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}} \\
    \\
 $
$
  \left[ {{H^ + }} \right] = \dfrac{{{K_a}}}{{\left[ {C{H_3}CO{O^ - }} \right]}} \times \left[ {C{H_3}COOH} \right] \\
    \\
 $
Substituting the values we get,
$
  \left[ {{H^ + }} \right] = \dfrac{{1.8 \times {{10}^{ - 5}}}}{{\left[ {0.20M} \right]}} \times \left[ {0.10\,M} \right] \\
    \\
 $
$
  \left[ {{H^ + }} \right] = 0.000009\,mol/L \\
    \\
 $
$\left[ {{H^ + }} \right] = 9 \times {10^{ - 6}}mol/L$
The concentration of $\left[ {{H^ + }} \right]$ ion is $9 \times {10^{ - 6}}\,mol/L$.

So, the correct answer is Option D .

Note:
We also remember that the dissociation reactions take place when one molecule is divided to form two smaller ones, leading to a decrease in energy. Dissociation reactions lead in the breakdown of a large molecule to form smaller products, giving them their second name: decomposition reactions. While some molecular compounds like water and acids produce electrolytic solutions, most dissociation reactions consist of ionic compounds in water, or aqueous solutions. When ionic compounds undergo dissociation, water molecules separate the ionic crystal. This takes place because of the attraction between the positive and negative ions in the crystal and the negative and positive polarity of water.