Question

Calculate the percentage ionization of 0.01 m acetic acid in 0.1 m hcl

Answer 1

Alpha = 1.8 × 10 ^ -5 / 0.1

= 1.8 ×10 ^ -4

Percentage = 1.8 × 10 ^ -4 × 100

= 0.018%

Hope it helps!

Answer 2

Answer:

The percentage ionisation of 0.01 m acetic acid on 0.1 m HCl is $0.018 \%$

Solution:

As we know that Ka (acid dissociation constant) of acetic acid is $1.8 \times 10^{-5}$

If we see the equation,  

$\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O}=\mathrm{CH}_{3} \mathrm{COO}(-)+\mathrm{H}_{3} \mathrm{O}(+)$

$K a\left(C H_{3} C O O H\right)=\frac{\left[C H_{3} C O O(-)\right]\left[H_{3} O(+)\right]}{\left[C H_{3} C O O H\right]}=\frac{\mathrm{x}(0.1+\mathrm{x})}{(0.1-\mathrm{x})}$                                        

If $\frac{c^{0}}{K a}>100$ then x will be negligible

So will get, $K a=\frac{0.1 x}{0.1}=x$

Hence,  

$\text { Alpha (percentage of ionization) }=\frac{x}{c^{\circ}}=\frac{K a}{c^{\circ}}=1.8 \times 10-\frac{5}{0.1}=1.8 \times 10-4=0.018 \%$