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Question: What is \(\left[ {{H^ + }} \right]\) in mol/L of a solution that is 0.20M in \(C{H_3}COONa\) and 0.1...

What is [H+]\left[ {{H^ + }} \right] in mol/L of a solution that is 0.20M in CH3COONaC{H_3}COONa and 0.10M in CH3COOHC{H_3}COOH? Ka{K_a} for CH3COOHC{H_3}COOH=1.8×10−51.8 \times {10^{ - 5}}
A. 3.5×10−43.5 \times {10^{ - 4}}
B. 1.1×10−51.1 \times {10^{ - 5}}
C. 1.8×10−51.8 \times {10^{ - 5}}
D. 9.0×10−69.0 \times {10^{ - 6}}

Explanation

Solution

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:
CO(g) + H2O(l)⇄CO2(g) + H2(g){\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 [H+]\left[ {{H^ + }} \right] by using the value of Ka{K_a} and concentrations of CH3COO−C{H_3}CO{O^ - } and concentration of CH3COOHC{H_3}COOH.

Formula used: We can calculate the concentration of hydronium ion using the equation,
[H+]=Ka[CH3COO−]×[CH3COOH]\left[ {{H^ + }} \right] = \dfrac{{{K_a}}}{{\left[ {C{H_3}CO{O^ - }} \right]}} \times \left[ {C{H_3}COOH} \right]
Here,
[H+]\left[ {{H^ + }} \right] is the concentration of hydronium ion
Ka{K_a} is the value of dissociation constant
[CH3COO−]\left[ {C{H_3}CO{O^ - }} \right] is the concentration of CH3COO−C{H_3}CO{O^ - }
[CH3COOH]\left[ {C{H_3}COOH} \right] is the concentration of CH3COOHC{H_3}COOH

Complete step by step answer:
Given data contains,
The concentration of CH3COOHC{H_3}COOH is 0.10 M0.10\,M.
The concentration of CH3COONaC{H_3}COONa is 0.20 M0.20\,M.
The value of Ka{K_a} for CH3COOHC{H_3}COOHis 1.8×10−51.8 \times {10^{ - 5}}.
We can write the ionization equation for CH3COOHC{H_3}COOH as,

We can write the ionization equation for CH3COONaC{H_3}COONa as,
CH3COONa→CH3COO−+Na+C{H_3}COONa\xrightarrow{{}}C{H_3}CO{O^ - } + N{a^ + }
We can that CH3COO−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.
Ka = [rate of products][rate of reactants] Ka=[CH3COO−][H+][CH3COOH]  {{\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 CH3COO−C{H_3}CO{O^ - }, CH3COOHC{H_3}COOH and Ka{K_a} to calculate the value of concentrations of [H+]\left[ {{H^ + }} \right] ions.
Ka=[CH3COO−][H+][CH3COOH]   {K_a} = \dfrac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}} \\\ \\\
[H+]=Ka[CH3COO−]×[CH3COOH]   \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,
[H+]=1.8×10−5[0.20M]×[0.10 M]   \left[ {{H^ + }} \right] = \dfrac{{1.8 \times {{10}^{ - 5}}}}{{\left[ {0.20M} \right]}} \times \left[ {0.10\,M} \right] \\\ \\\
[H+]=0.000009 mol/L   \left[ {{H^ + }} \right] = 0.000009\,mol/L \\\ \\\
[H+]=9×10−6mol/L\left[ {{H^ + }} \right] = 9 \times {10^{ - 6}}mol/L
The concentration of [H+]\left[ {{H^ + }} \right] ion is 9×10−6 mol/L9 \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.