Question

A $\mathbf{0}\mathbf{.}\mathbf{0450}\mathbf{M}$solution of HAis 0.60%dissociated. Calculate${\mathbf{pK}}_{\mathbf{a}}$for this acid.

Short answer

The ${\mathrm{pK}}_{\mathrm{a}}$ for HA is$=5.79$

Step-by-step solution

Define pKa

${\mathrm{pK}}_{\mathrm{a}}$is the number that describes the acidity of a particular molecule. The formula to find the ${\mathrm{pK}}_{\mathrm{a}}$ is

role="math" localid="1654857717351" ${\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{logK}}_{\mathrm{a}}$

Reaction of HA dissociation

$$\begin{gathered} \mathrm{HA}\stackrel{}{\rightleftharpoons }{\mathrm{H}}^{+}+{\mathrm{A}}^{-} \\ \left[\mathrm{HA}\right]=0.0450-\mathcal{x} \\ \left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{A}}^{-}\right]=\mathcal{x} \end{gathered}$$

Fraction of dissociation

Substituting the value of $\mathfrak{a}$and F in equation

$$\begin{gathered} \mathrm{a}=\frac{\mathrm{x}}{\mathrm{F}}=0.0060\Rightarrow \mathrm{x}=\mathrm{a}\times \mathrm{F} \\ \mathrm{x}=0.0060\times 0.0450 \end{gathered}$$

Calculating the value of  Ka

By using the formula${\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{x}}^2}{\mathrm{F}-\mathrm{x}}$

Substituting the value ofxand F we get

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{x}}^2}{\mathrm{F}-\mathrm{x}} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{{\left(0.0060\times 0.0450\right)}^2}{0.0450-\left(0.0060\times 0.0450\right)}=1.6\times {10}^{-6} \end{gathered}$$

 Step 5: Finding the value of pKa

Substituting the value of ${\mathrm{K}}_{\mathrm{a}}$in the formula role="math" localid="1654858632522" ${\mathrm{pK}}_{\mathrm{a}}$we get

${\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{logK}}_{\mathrm{a}}=-\log \left(1.6\times {10}^{-6}\right)=5.79$