In which of the following solutions will $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$be approximately equal to $\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]$:

$\begin{array}{l}{\mathbf{\text{(a)0.1MCH}}}_{\mathbf{\text{3}}}{\mathbf{\text{COOH}}}_{\mathbf{\text{;}}}\\ {\mathbf{\text{(b)1×10}}}^{\mathbf{\text{-J}}}{\mathbf{\text{MCH}}}_{\mathbf{\text{3}}}\mathbf{\text{COOH}}\\ {\mathbf{\text{(c)asolutioncontainingboth0.1MCH}}}_{\mathbf{\text{3}}}{\mathbf{\text{COOH}}}_{\mathbf{\text{;}}}{\mathbf{\text{and0.1MCH}}}_{\mathbf{\text{3}}}{\mathbf{\text{COOH}}}_{\mathbf{\text{;}}}\mathbf{\text{?Explain.}}\end{array}$

A dissociation reaction is a chemical reaction that breaks down a molecule into two or more components. A dissociation reaction's general formula is as follows: A + B = AB

(a)

First, we'll write the given dissociation response.

${\text{CH}}_{\text{3}}{\text{COOH+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}$

As we can see, the reaction created an equal amount of ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}$and ${\text{H}}_3{\text{O}}^{+}$because it produced 1 mole of each. so$\left[{\text{CH}}_3{\text{COO}}^{-}\right]\approx \left[{\text{H}}_3{\text{O}}^{+}\right]$

Therefore, the dissociation reaction is approximately equal.

(b)

It will be identical to (a). The moles of the product generated will not be affected by the concentration. So $\left[{\text{CH}}_3{\text{COO}}^{-}\right]\approx \left[{\text{H}}_3{\text{O}}^{+}\right]$$\left[{\text{CH}}_3{\text{COO}}^{-}\right]\approx \left[{\text{H}}_3{\text{O}}^{+}\right]$

Hence, the dissociation reaction is approximately equal.

(c)

However, there is the presence of${\text{CH}}_{\text{3}}\text{COONa}$, ${\text{CH}}_{\text{3}}\text{COONa}$ which is water-soluble salt. As a result, it will totally dissociate in water and generate an equivalent amount of ${\text{CH}}_{\text{3}}\text{COO}$.

${\text{CH}}_3\text{COONa}\to {\text{CH}}_3{\text{COO}}^{-}+{\text{Na}}^{+}$

The concentration of will now rise by$0.1\text{M}$ a factor of $\left[{\text{H}}_3{\text{O}}^{+}\right]$. Because of Le-Chatelaine’s principle, it may devour some of the with its existence. Now that we have more products, the reaction will be driven in the direction of the reactant. So$\left[{\text{CH}}_3{\text{COO}}^{-}\right]\ne \left[{\text{H}}_3{\text{O}}^{+}\right]$

As a result, the dissociation reaction is not approximately equal.