The ionization constant of acetic acid is $1.74 \times 10^{-5}$. Calculate the degree of dissociation of acetic acid in its $0.05\, M$ solution. Calculate the concentration of acetate ion in the solution and its $pH$.
Method $1$
$(1)$ $CH _{3} COOH \longleftrightarrow CH _{3} COO ^{-}+ H ^{-+}$ $K_{a}=1.74 \times 10^{-5}$
$(2)$ $H _{2} O + H _{2} O \longleftrightarrow H _{3} O ^{+}+ OH ^{-}$ ${K_w} = 1.0 \times {10^{ - 14}}$
Since $K a\,>\,>\,K_{w,}$
$CH _{3} COOH + H _{2} O \longleftrightarrow CH _{3} COO ^{-}+ H _{3} O ^{+}$
$C_{i}=$ $0.05$ $0$ $0$
$0.05-.05 \alpha$ $0.05 \alpha$ $0.05 \alpha$
$K_{a}=\frac{(.05 \alpha)(.05 \alpha)}{(.05-0.05 \alpha)}$
$=\frac{(.05 \alpha)(0.05 \alpha)}{.05(1-\alpha)}$
$=\frac{.05 \alpha^{2}}{1-\alpha}$
$1.74 \times 10^{-5}=\frac{0.05 \alpha^{2}}{1-\alpha}$
$1.74 \times 10^{-5}-1.74 \times 10^{-5} \alpha=0.05 \alpha^{2}$
$0.05 \alpha^{2}+1.74 \times 10^{-5} \alpha-1.74 \times 10^{-5}$
$ D =b^{2}-4 a c $
$=\left(1.74 \times 10^{-5}\right)^{2}-4(.05)\left(1.74 \times 10^{-5}\right) $
$=3.02 \times 10^{-25}+.348 \times 10^{-5} $
$\alpha =\sqrt{\frac{K_{a}}{c}} $
$\alpha=\sqrt{\frac{1.74 \times 10^{-5}}{.05}}$
$=\sqrt{\frac{34.8 \times 10^{-5} \times 10}{10}}$
$=\sqrt{3.48 \times 10^{-6}}$
$= CH _{3} COOH \longleftrightarrow CH _{3} COO ^{-}+ H ^{+}$
$\alpha 1.86 \times 10^{-3}$
$\left[ CH _{3} COO ^{-}\right]=0.05 \times 1.86 \times 10^{-3}$
$=\frac{0.93 \times 10^{-3}}{1000} $
$=.000093$
Method $2$
Degree of dissociation,
$ \alpha =\sqrt{\frac{K_{a}}{c}} $
$c =0.05 M $
$ K_{a} =1.74 \times 10^{-5} $
Then, $\alpha=\sqrt{\frac{1.74 \times 10^{-5}}{.05}}$
$\alpha=\sqrt{34.8 \times 10^{-5}}$
$\alpha=\sqrt{3.48} \times 10^{-4}$
$\alpha=1.8610^{-2}$
$CH _{3} COOH \longleftrightarrow CH _{3} COO ^{-}+ H ^{+}$
Thus, concentration of $CH _{3} COO -= c.a$
$=.05 \times 1.86 \times 10^{-2}$
$=.093 \times 10^{-2}$
$=.00093 \,M$
Since $\left[ oAc ^{-}\right]=\left[ H ^{+}\right]$
$\left[ H ^{+}\right]=.00093=.093 \times 10^{-2}$
$ pH =-\log \left[ H ^{+}\right] $
$=-\log \left(.093 \times 10^{-2}\right) $
$\therefore pH =3.03 $
Hence, the concentration of acetate ion in the solution is $0.00093 \,M$ and its $Ph$ is $3.03$
Assuming that the degree of hydrolysis is small, the $pH$ of $0.1\, M$ solution of sodium acetate $(K_a\, = 1.0\times10^{- 5})$ will be
Determine the degree of ionization and $pH$ of a $0.05 \,M$ of ammonia solution. The ionization constant of ammonia can be taken from Table $7.7 .$ Also, calculate the ionization constant of the conjugate acid of ammonia.
What is the percent ionization $(\alpha)$ of a $0.01\, M\, HA$ solution ? .......$\%$ $(K_a = 10^{-6})$
When $100 \ mL$ of $1.0 \ M \ HCl$ was mixed with $100 \ mL$ of $1.0 \ M \ NaOH$ in an insulated beaker at constant pressure, a temperature increase of $5.7^{\circ} C$ was measured for the beaker and its contents (Expt. $1$). Because the enthalpy of neutralization of a strong acid with a strong base is a constant $\left(-57.0 \ kJ \ mol ^{-1}\right)$, this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. $2$), $100 \ mL$ of $2.0 \ M$ acetic acid $\left(K_a=2.0 \times 10^{-5}\right)$ was mixed with $100 \ mL$ of $1.0 M \ NaOH$ (under identical conditions to Expt. $1$) where a temperature rise of $5.6^{\circ} C$ was measured.
(Consider heat capacity of all solutions as $4.2 J g ^{-1} K ^{-1}$ and density of all solutions as $1.0 \ g mL ^{-1}$ )
$1.$ Enthalpy of dissociation (in $kJ mol ^{-1}$ ) of acetic acid obtained from the Expt. $2$ is
$(A)$ $1.0$ $(B)$ $10.0$ $(C)$ $24.5$ $(D)$ $51.4$
$2.$ The $pH$ of the solution after Expt. $2$ is
$(A)$ $2.8$ $(B)$ $4.7$ $(C)$ $5.0$ $(D)$ $7.0$
Give the answer question $1$ and $2.$
Calculate the $pH$ of $0.08\, M$ solution of hypochlorous acid, $HOCl$. The ionization constant of the acid is $2.5 \times 10^{-5}$ Determine the percent dissociation of $HOCl.$