Question

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

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

Short answer

1. The dissociation reaction is approximately equal.
2. The dissociation reaction is approximately equal.
3. The dissociation reaction is not approximately equal.

Step-by-step solution

Explanation of the concept

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

Finding the dissociation reaction for part (a)

(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.

Finding the dissociation reaction for part (b)

(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.

Finding the dissociation reaction for part (c)

(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.