The base B has ${}^{{\mathbf{pK}}_{\mathbf{b}}}$= 5.00.

(a) What is the value of ${}^{{\mathbf{pK}}_{\mathbf{a}}}$for the acid ${}^{{\mathbf{BH}}^{\mathbf{+}}}$?

(b) At what pH is [${}^{{\mathbf{BH}}^{\mathbf{+}}}$] = [B]?

(c) Which is the principal species at pH 7.00: B or ${}^{{\mathbf{BH}}^{\mathbf{+}}}$?

(d) What is the quotient [B] / [${}^{{\mathbf{BH}}^{\mathbf{+}}}$] at pH 12.00?

In this task we have a base with${}^{{\mathbf{pK}}_{\mathbf{b}}}$= 5.00, we need to determine the following:

a) the value of ${}^{{\mathbf{pK}}_{\mathbf{a}}}$for the acid ${}^{{\mathbf{BH}}^{\mathbf{+}}}$

b) the pH when${}^{\mathbf{\left[}\mathbf{B}\mathbf{\right]}=\mathbf{\left[}{\mathbf{BH}}^{\mathbf{+}}\mathbf{\right]}}$

c) the principal species at pH = 7.00
d) the quotient ${}^{\mathbf{\left[}\mathbf{B}\mathbf{\right]}\mathbf{/}\mathbf{\left[}{\mathbf{BH}}^{\mathbf{+}}\mathbf{\right]}}$at pH = 12.00

(a) we will calculate the ${}^{{\mathbf{pK}}_{\mathbf{a}}}$by the following:

$$\begin{gathered} {\mathbf{pK}}_{\mathbf{a}}\mathbf{+}{\mathbf{pK}}_{\mathbf{b}}\mathbf{=}\mathbf{14} \\ {\mathbf{pK}}_{\mathbf{a}}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{=}\mathbf{14}\mathbf{.}\mathbf{00} \\ {\mathbf{pK}}_{\mathbf{a}}\mathbf{}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{00} \end{gathered}$$

(b) Calculate the pOH when ${}^{\left[{\mathrm{BH}}^{+}\right]\mathbf{=}\left[B\right]}$via Henderson-Hasselbach equation;

$$\begin{gathered} \mathbf{pOH}\mathbf{}\mathbf{=}{\mathbf{pK}}_{\mathbf{b}}\mathbf{+}\mathbf{log}\mathbf{}\left(\left[B\right]/\left[{\mathrm{BH}}^{+}\right]\right) \\ \mathbf{pOH}\mathbf{}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{+}\mathbf{log}\left(1\right) \\ \mathbf{pOH}\mathbf{}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{+}\mathbf{0} \\ \mathbf{pOH}\mathbf{}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00} \end{gathered}$$

Then use the following to get pH of the solution:

role="math" localid="1654771588201" $$\begin{gathered} \mathbf{pH}\mathbf{+}\mathbf{pOH}\mathbf{=}\mathbf{14}\mathbf{.}\mathbf{00} \\ \mathbf{pH}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{=}\mathbf{14}\mathbf{.}\mathbf{00} \\ \mathbf{pH}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{00} \end{gathered}$$

c) the principal species at pH=7.00, we need to use the previous equation and values of ${}^{{\mathbf{pK}}_{\mathbf{a}}}$from step 2:

$$\begin{gathered} \mathbf{pH}\mathbf{}\mathbf{=}{\mathbf{pK}}_a\mathbf{+}\mathbf{log}\mathbf{}\mathbf{(}\mathbf{\left[}\mathbf{B}\mathbf{\right]}\mathbf{/}\mathbf{[}{\mathbf{BH}}^{\mathbf{+}}\mathbf{}\mathbf{]}\mathbf{)} \\ \mathbf{7}\mathbf{.}\mathbf{00}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{00}\mathbf{+}\mathbf{log}\mathbf{}\mathbf{(}\mathbf{\left[}\mathbf{B}\mathbf{\right]}\mathbf{/}\mathbf{[}{\mathbf{BH}}^{\mathbf{+}}\mathbf{}\mathbf{]}\mathbf{)} \\ \mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{=}\mathbf{log}\mathbf{}\mathbf{(}\mathbf{\left[}\mathbf{B}\mathbf{\right]}\mathbf{/}\mathbf{[}{\mathbf{BH}}^{\mathbf{+}}\mathbf{}\mathbf{]}\mathbf{)} \\ {\mathbf{10}}^{\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}}\mathbf{=}\mathbf{log}\mathbf{}\mathbf{\left[}\mathbf{B}\mathbf{\right]}\mathbf{/}\mathbf{[}{\mathbf{BH}}^{\mathbf{+}}\mathbf{}\mathbf{]} \\ \mathbf{0}\mathbf{.}\mathbf{01}\mathbf{=}\mathbf{log}\mathbf{}\mathbf{\left[}\mathbf{B}\mathbf{\right]}\mathbf{/}\mathbf{[}{\mathbf{BH}}^{\mathbf{+}}\mathbf{}\mathbf{]} \end{gathered}$$

Considering the quotient ${}^{{\mathbf{BH}}^{\mathbf{+}}}$is the principal species

(d) the quotient ${}^{\mathbf{}\mathbf{\left[}\mathbf{B}\mathbf{\right]}\mathbf{/}\mathbf{[}{\mathbf{BH}}^{\mathbf{+}}\mathbf{}\mathbf{]}}$at pH = 12.00, also use the previous equation and value of ${}^{{\mathbf{pK}}_{\mathbf{a}}}$from step 2;

$$\begin{gathered} \mathbf{pH}\mathbf{}\mathbf{=}{\mathbf{pK}}_{\mathbf{a}}\mathbf{+}\mathbf{log}\mathbf{}\mathbf{(}\mathbf{\left[}\mathbf{B}\mathbf{\right]}\mathbf{/}\mathbf{[}{\mathbf{BH}}^{\mathbf{+}}\mathbf{}\mathbf{]}\mathbf{)} \\ \mathbf{12}\mathbf{.}\mathbf{00}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{00}\mathbf{+}\mathbf{log}\mathbf{}\mathbf{(}\mathbf{\left[}\mathbf{B}\mathbf{\right]}\mathbf{/}\mathbf{[}{\mathbf{BH}}^{\mathbf{+}}\mathbf{}\mathbf{]}\mathbf{)} \\ \mathbf{3}\mathbf{.}\mathbf{00}\mathbf{=}\mathbf{log}\mathbf{}\mathbf{(}\mathbf{\left[}\mathbf{B}\mathbf{\right]}\mathbf{/}\mathbf{[}{\mathbf{BH}}^{\mathbf{+}}\mathbf{}\mathbf{]}\mathbf{)} \\ {\mathbf{10}}^{\mathbf{3}\mathbf{.}\mathbf{00}}\mathbf{=}\mathbf{log}\mathbf{}\mathbf{\left[}\mathbf{B}\mathbf{\right]}\mathbf{/}\mathbf{[}{\mathbf{BH}}^{\mathbf{+}}\mathbf{}\mathbf{]} \\ \mathbf{1000}\mathbf{=}\mathbf{}\mathbf{\left[}\mathbf{B}\mathbf{\right]}\mathbf{/}\mathbf{[}{\mathbf{BH}}^{\mathbf{+}}\mathbf{}\mathbf{]} \end{gathered}$$