11-18. Calculate the ${}^{\mathrm{pH}}$of a solution made by mixing${}^{50.00\mathrm{mL}}$of ${}^{\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{}}$$\mathbf{M}$$\mathbf{NaCN}$with

(a)${}^{\mathbf{4}\mathbf{.}\mathbf{20}\mathbf{mL}}$Of ${}^{\mathbf{4}\mathbf{.}\mathbf{38}}$$\mathbf{M}$

(b)${}^{\mathbf{4}\mathbf{.}\mathbf{20}\mathbf{mL}}$Of ${}^{\mathbf{4}\mathbf{.}\mathbf{38}}$$\mathbf{M}$

(c) What is the ${}^{\mathbf{pH}}$at the equivalence point with${}^{\mathbf{4}\mathbf{.}\mathbf{38}}$$\mathbf{M}$${}^{{\mathbf{HCIO}}_{\mathbf{4}}}$?

A polyprotic acid contributes protons in the same way as an Arrhenius acid donates a proton ${}^{\left(H+\right)}$. A polyprotic acid, on the other hand, differs from a monoprotic acid in that it contains more than one acidic ${}^{\mathbf{H}\mathbf{+}}$and can thus contribute to numerous protons. It does not entirely dissociate as a weak polyprotic acid. Some examples of weak polyprotic acids are as follows:

• ${\mathbf{H}}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{4}}$ is a triphosphate acid.

• ${\mathbf{H}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$as a diprotic acid

• ${\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{3}}$is a diprotic acid

a)

In the a) task we have a titration of a weak base ${}^{\left(\mathrm{NaCN}\right)}$with a strong acid localid="1655014522770" ${}^{(\mathrm{HCIO}4)}$

To make the calculations easier we can find the equivalent point. That is the volume of acid needed to neutralize the base:

localid="1655016807384" $\mathbf{V}\mathbf{\left(}{\mathbf{HCIO}}_{\mathbf{4}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{c}\mathbf{\left(}\mathbf{NaCN}\mathbf{\right)}\mathbf{.}\mathbf{}\mathbf{V}\mathbf{\left(}\mathbf{NaCN}\mathbf{\right)}}{\mathbf{c}\mathbf{(}\mathbf{HCIO}\mathbf{}\mathbf{4}\mathbf{)}}$

$\mathbf{V}\mathbf{\left(}{\mathbf{HCIO}}_{\mathbf{4}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{mL}\mathbf{}\mathbf{.}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{M}}{\mathbf{0}\mathbf{.}\mathbf{438}\mathbf{M}}$

$\mathbf{V}\mathbf{\left(}{\mathbf{HCIO}}_{\mathbf{4}}\mathbf{\right)}\mathbf{=}\mathbf{}\mathbf{11}\mathbf{.}\mathbf{42}\mathbf{mL}$

Since the volume ${}^{{\mathbf{HCIO}}_{\mathbf{4}}}$needed to reach the equivalence point is ${}^{\mathbf{11}\mathbf{.}\mathbf{42}\mathbf{mL}}$, when we add ${}^{\mathbf{4}\mathbf{.}\mathbf{20}\mathbf{mL}}$it, the remaining volume of ${}^{\mathbf{B}}$is ${}^{\mathbf{7}\mathbf{.}\mathbf{22}\mathbf{mL}}$and the volume of${}^{{\mathbf{BH}}^{\mathbf{+}}}$is ${}^{\mathbf{4}\mathbf{.}\mathbf{20}\mathbf{mL}}$since its equivalent to the volume of the added acid.

The${}^{{\mathbf{K}}_{\mathbf{b}}}$value ${}^{\mathbf{NaCN}}$can be found in the appendix ${}^{\mathbf{G}}$under HCN.

Now we can calculate the${}^{\mathbf{pH}}$by using the Henderson-Hasselbalch equation:

At ${}^{\mathbf{4}\mathbf{.}\mathbf{20}\mathbf{mL}}$:

localid="1655016033745" $\mathbf{pOH}\mathbf{}\mathbf{=}{\mathbf{pK}}_{\mathbf{b}}\mathbf{+}\mathbf{log}\frac{\left[{\mathrm{BH}}^{+}\right]}{\left[B\right]}$

$\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{79}\mathbf{+}\mathbf{log}\frac{\mathbf{4}\mathbf{.}\mathbf{20}\mathbf{/}\mathbf{11}\mathbf{.}\mathbf{42}}{\mathbf{7}\mathbf{.}\mathbf{22}\mathbf{/}\mathbf{11}\mathbf{.}\mathbf{42}}$

$\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{55}\mathbf{pH}$

$\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{14}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{43}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{45}$

b)

In the task b) the volume of ${}^{{\mathbf{HCIO}}_{\mathbf{4}}}$is ${}^{\mathbf{11}\mathbf{.}\mathbf{82}\mathbf{}\mathbf{mL}}$and that means it is after the equivalence point so we can calculate the ${}^{\mathbf{pH}}$by the excess of ${}^{{\mathbf{H}}^{\mathbf{+}}}$:

At ${}^{\mathbf{11}\mathbf{.}\mathbf{82}\mathbf{}\mathbf{mL}}$:

$\left[H^{+}\right]\mathbf{=}\mathbf{c}\mathbf{}\left(\mathrm{initial}{\mathrm{HCIO}}_4\right)\mathbf{}\frac{\mathbf{V}\left(\mathrm{excess}{\mathrm{HCIO}}_4\right)}{\mathbf{V}\left(\mathrm{solution}\right)}$

$\mathbf{\left[}{\mathbf{H}}^{\mathbf{+}}\mathbf{\right]}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{438}\mathbf{.}\mathbf{}\frac{\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{mL}}{\mathbf{50}\mathbf{mL}\mathbf{+}\mathbf{11}\mathbf{.}\mathbf{82}\mathbf{mL}}$

$\mathbf{\left[}{\mathbf{H}}^{\mathbf{+}}\mathbf{\right]}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{83}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{M}$

$\mathbf{pH}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{55}$

c):

To calculate ${}^{\mathbf{pH}}$at the equivalence point ${}^{\left(11.42\mathrm{mL}\right)}$we need to write the reaction of the weak acid with water

${\mathbf{BH}}^{\mathbf{+}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{B}\mathbf{+}{\mathbf{H}}^{\mathbf{+}}$

The concentrations are:

$\left[{\mathrm{BH}}^{+}\right]\mathbf{=}{\mathbf{F}}^{\mathbf{\text{'}}}\mathbf{-}\mathbf{x}$

$\mathbf{\left[}\mathbf{B}\mathbf{\right]}\mathbf{=}\mathbf{x}$

$\mathbf{\left[}{\mathbf{H}}^{\mathbf{+}}\mathbf{\right]}\mathbf{=}\mathbf{x}$

The only thing we need to do before putting the concentrations in the weak acid equation is to calculate the formal concentration, ${}^{{\mathbf{F}}^{\mathbf{\text{'}}}}$:

${\mathbf{F}}^{\mathbf{\text{'}}}\mathbf{}\mathbf{=}\mathbf{concentration}\mathbf{}\mathbf{of}\mathbf{}\mathbf{initial}\mathbf{}\mathbf{B}\mathbf{}\frac{\mathbf{initial}\mathbf{}\mathbf{volume}\mathbf{}\mathbf{of}\mathbf{}\mathbf{B}}{\mathbf{total}\mathbf{}\mathbf{volume}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{solution}\mathbf{}}$

${\mathbf{F}}^{\mathbf{\text{'}}\mathbf{}}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{M}\mathbf{.}\frac{\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{mL}}{\mathbf{61}\mathbf{.}\mathbf{42}\mathbf{mL}}$

${\mathbf{F}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0814}\mathbf{M}$

Now we can insert it into the equation for the equilibrium of the weak acid:

${\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{\left[B\right]\left[H^{+}\right]}{\left[{\mathrm{BH}}^{+}\right]}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}\mathbf{0814}\mathbf{-}\mathbf{x}}$

By rearranging the equation we get a quadratic equation:

${\mathit{x}}^{\mathbf{2}\mathbf{+}}\mathbf{6}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{\times }\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{047}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{}\mathbf{=}\mathbf{0}$

By solving the quadratic equation we get the ${}^{\mathbf{ph}}$at ${}^{\mathbf{31}\mathbf{.}\mathbf{9}\mathbf{mL}}$:

$\mathbf{x}\mathbf{=}\left[H^{+}\right]\mathbf{=}\mathbf{}\mathbf{7}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{M}$

$\mathbf{pH}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{15}$