The following scenes represent three weak acids HA

(a) Rank the acids in order of increasing Ka.

(b) Rank the acids in order of increasing${\mathbf{pK}}_{\mathbf{a}}.$

c) Rank the conjugate bases in order of increasing ${\mathbf{pK}}_{\mathbf{b}}$.

(d) What is the percent dissociation of HX?

(e) If equimolar amounts of the sodium salts of the acids (NaX, NaY, and NaZ) were dissolved in water, which solution would have the highest pOH? The lowest pH?

The chemical equation of the reaction is:

$\mathrm{HA}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}\left(\mathrm{aq}\right)+{\mathrm{A}}^{-}\left(\mathrm{aq}\right)$

The expression for Ka is:

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}$

Rank acids based on increasing${\text{K}}_{\text{a}}$ meaning that the${\text{K}}_{\text{a}}$. The acid is also large.

$$\begin{gathered} {\text{K}}_a \text{for} \text{HX:}\frac{2\times 2}{6}=0.667 \\ {\text{K}}_a \text{for} \text{HY:}\frac{6\times 6}{2}=18 \\ {\text{K}}_a\text{for HZ:}\frac{4\times 4}{4}=4 \end{gathered}$$

Since the ratio is $\mathrm{HX}<\mathrm{HZ}<\mathrm{HY}$, this will also follow the order of increasing${\text{K}}_{\text{a}}$

Therefore, the order of increasing${\text{K}}_{\text{a}}$ of the following acids is $HX<HZ<HY$.

Rank acids based on increasing ${\text{pK}}_{\text{a}}$

Note that the ${\text{pKa = - logK}}_{\text{a}}{\text{.AsK}}_{\text{a}}$increases, ${\text{pK}}_{\text{a}}$decreases.

Therefore, the order of increasing ${\text{pK}}_{\text{a}}$is just the inverse of the order of increasing ${\text{K}}_{\text{a}}$.

Rank conjugate bases based on increasing${\text{pK}}_{\text{b}}$

Note that, a stronger acid will yield a weaker conjugate base. The conjugate base of a weaker acid is stronger than the conjugate base of a stronger acid. Therefore, the rank of increasing order of base strength or${\text{K}}_{\text{b}}$ of the conjugate bases, is just the inverse of the order of increasing${\text{K}}_{\text{a}}$ of its respective acids.$Y^{-}<Z^{-}<X^{-}$. Now, note that the${\mathrm{pK}}_{\mathrm{b}}=-{\mathrm{logK}}_{\mathrm{b}}$. As${\text{K}}_{\text{b}}$ increases${\text{pK}}_{\text{b}}$ decreases.

Therefore, the order of increasing${\text{pK}}_{\text{b}}$ is just the inverse of the order of increasing $Y^{-}<Z^{-}<X^{-}$.

%dissociation of

To solve for % dissociation, use the formula below.

$\%\mathrm{dissociation}=\frac{{\left[\mathrm{HA}\right]}_{\mathrm{dissoc}}}{{\left[\mathrm{HA}\right]}_{int}}\times 100\%$

First, let us solve for the $\mathrm{HA}{]}_{int}\mathrm{of}\mathrm{HX}.[\mathrm{HA}{]}_{int}=[\mathrm{HA}{]}_{\mathrm{dissoc}}+\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$in the solution.

role="math" localid="1663275818262" $[HX{]}_{int}=2+6=8$

Then solve for the % dissociation.

$\%\text{dissociation}=\frac{2}{8}\times 100\%=25\%$

First, NaA will dissolve in water.

$\text{NaA}\to {\text{Na}}^{\text{+}}{\text{+A}}^{\text{-}}$

Then $A^{-}$will react with water.

${\text{A}}^{-}+{\text{H}}_2\text{O}\rightleftharpoons \text{HA}+{\text{OH}}^{-}$

For the highest pOH, it must be the solution with the lowest${\text{K}}_{\text{b}}$ since it did not dissociate high amounts of${\text{OH}}^{\text{-}}$ compared to the others. The conjugate base with the Lowest${\text{K}}_{\text{b}}$ based on (c) is ${\text{Y}}^{\text{-}}$.Therefore, the sodium salt with the highest pOH is NaY. Similarly, the relationship between pH and pOH is inverse.

Therefore, the sodium salt with the lowest pH is NaY