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• To determine the pH = pKa + log([A]/[HA]) of a buffer, utilize the Henderson-Hasselbalch equation.
• The equilibrium concentrations of the conjugate acid-base pair utilized to form the buffer solution are [HA] and [A] in this equation.
• The pH of the solution is equal to the acid's pKa when [HA] = [A] . Jay came up with the idea.

A solution prepared from 10mL of 0.1M cacodylic acid (HA) and 10mL of 0.08MNaOH To this mixture was added 1mL of $1.27脳{10}^{-6}M$morphine (B) . We need to calculate the fraction of morphine present in the form $B{H}^{+}$.

First calculate the moles of:

a) 10mL of 0.1M cacodylic acid (HA)

$$\begin{gathered} \mathrm{n}\left(\mathrm{HA}\right)=\mathrm{c}脳\mathrm{Vn}\left(\mathrm{HA}\right) \\ =0.1\mathrm{M}脳10\mathrm{mLn}\left(\mathrm{HA}\right) \\ =1\mathrm{mmol} \end{gathered}$$

b) 10mL of 0.08MNaOH

$$\begin{gathered} \mathrm{n}\left({\mathrm{OH}}^{-}\right)=\mathrm{c}脳\mathrm{Vn}\left({\mathrm{OH}}^{-}\right) \\ =0.08\mathrm{M}脳10\mathrm{mLn}\left({\mathrm{OH}}^{-}\right) \\ =0.8\mathrm{mmol} \end{gathered}$$

Consider the reaction

$HA+O{H}^{-}鈬�A^{-}+H_{2}O$

Initial mmol:

$$\begin{gathered} \left[\mathrm{HA}\right]=1\mathrm{mmol} \\ \left[{\mathrm{OH}}^{-}\right]=0.8\mathrm{mmol} \\ \left[{\mathrm{A}}^{-}\right]=0.8\mathrm{mmol} \end{gathered}$$

Final mmol:

$$\begin{gathered} \left[\mathrm{HA}\right]=0.2\mathrm{mmol} \\ \left[{\mathrm{A}}^{-}\right]=0.8\mathrm{mmol} \end{gathered}$$

Determine the value of ${\mathrm{pK}}_{\mathrm{a}}$for cacodylic acid (HA) by the following:

$$\begin{gathered} {\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{logK}}_{\mathrm{a}} \\ {\mathrm{pK}}_{\mathrm{a}}=-\log \left(6.4脳{10}^{-7}\right) \\ {\mathrm{pk}}_{\mathrm{a}}=6.19 \end{gathered}$$

Next calculate the pH of cacodylic acid (HA) using the Henderson-Hasselbach equation

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \left(\left[{\mathrm{A}}^{-}\right]/\left[\mathrm{HA}\right]\right) \\ \mathrm{pH}=6.19+\log \left(0.8/0.2\right) \\ \mathrm{pH}=6.79 \end{gathered}$$

Then calculate the $K_a$for morphine (B) :

$$\begin{gathered} K_a=-K_w/K_b \\ {K}_a={10}^{-14}/\left(1.6脳{10}^{-6}\right) \\ {K}_a=6.25脳{10}^{-9} \end{gathered}$$

which gives us the $p{K}_a$for morphine (B) :

$$\begin{gathered} p{K}_a=-\log K_a \\ p{K}_a=-\log \left(6.25脳{10}^{-9}\right) \\ p{K}_a=8.20 \end{gathered}$$

Determine the quotient by the Henderson-Hasselbach equation:

$$\begin{gathered} pH=p{K}_a+\log \left(\left[B\right]/\left[B{H}^{+}\right]\right) \\ 6.79=8.20+\log \left(\left[B\right]/\left[B{H}^{+}\right]\right) \\ -1.41=\log \left(\left[B\right]/\left[B{H}^{+}\right]\right) \\ {10}^{-1.41}=\left(\left[B\right]/\left[B{H}^{+}\right]\right) \\ 0.039=\left(\left[B\right]/\left[B{H}^{+}\right]\right) \end{gathered}$$

Next write the fraction for morphine (B) in the form of $B{H}^{+}$:

$$\begin{gathered} 伪=\frac{\left[B{H}^{+}\right]}{\left[B\right]+\left[B{H}^{+}\right]} \\ 伪=\frac{\left[B{H}^{+}\right]}{0.039\left[B{H}^{+}\right]+\left[B{H}^{+}\right]} \\ 伪=0.96 \end{gathered}$$

Therefore, the fraction of morphine B is 0.96