Question

$0.6mL$ of acetic acid $C{H}_{3}COOH$, having density $1.06gm{L}^{-1}$, is dissolved in $1$ litre of water. The depression in freezing point observed for this strength of acid was ${0.0205}^{\circ }C$. Calculate the van’t Hoff factor and the dissociation constant of acid.

Answer 1

Given:
Volume of acetic acid $=0.6mL$,
Density of acetic acid0 $=1.06gm{L}^{-1}$,
Volume of water (solvent) $=1L$,
Depression in freezing point =${0.0205}^{\circ }C\text{or}0.0205K$

Moles of acetic acid

We know, density $=\frac{\text{mass}}{\text{volume}}$

So, mass of acetic acid (solute)$=0.6mL\times 1.06gm{L}^{-1}=0.636g$

Number of moles of acetic acid $=\frac{\text{mass of acetic acid}}{\text{molar mass of acetic acid}}$
$=\frac{0.636g}{60gmo{l}^{-1}}=0.0106mol=n$

Molality of solution

We know, density of water$=1gm{L}^{-1}\text{or}kg{L}^{-1}$

Mass of water (solvent) $=1kg$

Molality(m) $=\frac{\text{mole of acetic acid}}{\text{molar mass of acetic acid}\times \text{mass of solvent(kg)}}$

Molality(m) $=\frac{0.0106mol}{1kg}=0.00106molk{g}^{-1}$

Calculated depression in freezing point

We know that $\Delta T_f=K_f\times m$
$\Delta T_f=1.86\times Kkgmol6-1\times 0.0106molk{g}^{-1}=0.0197K$

van't Hoff factor

van't Hoff factor $\left(i\right)$
$\frac{\text{observed freezing point depression}}{\text{Calculated freezing point depression}}$

$\frac{0.0205K}{0.0197K}=1.041$

Acetic acid is a weak electrolyte and will dissociate into two ions: acetate and hydrogen ions.
Let $x$ be the degree of dissociation of acetic acid. Then we would have $n(1-x)$ moles of undissociated acetic acid, $nx$ moles of $C{H}_{3}CO{O}^{-}\text{and}nx$ moles of $H^{+}$ ions.

Now consider the following equilibrium for the acid:
$C{H}_{3}COOH\rightleftharpoons H^{+}+C{H}_{3}CO{O}^{-}$

$\begin{array}{cccc}\text{At t}=0& 0& 0& 0\\ \text{At t}=t_{eq}& n(1-x)& nx& nx\end{array}$

Thus, total moles of particles are:
$n(1-x+x+x)=n(1+x)$

$i=\frac{\text{total moles at equilibrium}}{\text{intial moles}}=\frac{n(1+x)}{n}$
$=1+x=1.041$
Thus, degree of dissociation of acetic acid
$\left(x\right)=1.041-1.000=0.041$

Then,
$\left[C{H}_{3}COOH\right]=n(1-x)$
$=0.0106(1-0.041)M$
$\left[C{H}_{3}CO{O}^{-}\right]=nx$
$=0.0106\times 0.041M$

and $\left[H^{+}\right]=nx$
$0.0106\times 0.041M$
Dissociation constant of acid

van't Hoff Factor $\left(i\right)=\frac{\text{Observed freezing point}}{\text{Calculated freezing point}}=\frac{0.0205K}{0.0197K}$
$=1.041$

Acetic acid is a weak electrolyte and
will dissociate into two ions: acetate and
hydrogen ions
$\left[H^{+}\right]=nx=0.0106\times 0.041$

$K_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}$

$\frac{0.0106\times 0.041\times 0.0106\times 0.041}{0.0106(1.00-0.041)}$
$=1.86\times {10}^{-5}$