calculate the degree of ionization of 0.05M acetic acid if Pka =4.14 how is the degree of dissociation affected what its solution is also 0.01 M

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Solution

$C = 0.05 M p K_{a} = 4.74 - \log K_{a} = 4.74 K_{a} = A . l g (- 4.74) = 1.82 \times 10^{- 5} K_{a} = α^{2} C α = \sqrt{\frac{K_{a}}{C}} = \sqrt{\frac{1.82 \times 10^{- 5}}{0.05}} = 1.908 \times 10^{- 2} W h e n 0.01 M H C l i s t a k e n, x b e t h e a m o u n t o f a c e t i c a c i d d i s s o c i a t e d a f t e r t h e a d d i t i o n o f H C l . C H_{3} C O O H \leftrightarrow H^{+} + C H_{3} C O O^{-} I n i t i a l c o n c e n t r a t i o n 0.05 0 0 A f t e r d i s s o c i a t i o n 0.05 - x 0.01 + x x A s t h e d i s s o c i a t i o n i s v e r y s m a l l 0.05 - x \approx 0.05 a n d 0.01 + x \approx 0.01 K_{a} = \frac{[C H_{3} C O O^{-}] [H^{+}]}{[C H_{3} C O O H]} = \frac{0.01 \times x}{0.05} 1.82 \times 10^{- 5} = \frac{0.01 \times x}{0.05} x = 1.82 \times 10^{- 3} \times 0.05 D e g r e e o f d i s s o c i a t i o n = \frac{A m o u n t o f a c i d d i s s o c i a t e d}{A m o u n t o f a c i d t a k e n} = \frac{1.82 \times 10^{- 3} \times 0.05}{0.05} = 1.82 \times 10^{- 3}$

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