Question

Help?

Answer 1

The Ionization constant if HF is 3.2*10^-4.calculate degree of dissociation of HF in its 0.02M solution

(A) Here we use the Formula degree of dissociation = sqrt (ka/c)


(2) The inonization constant of acetic acid is 7.74 *10^-5. Calculate Degree of Dissocaition
in its 0.05M solution

(A) Here we use the formula degree of dissociation = sqrt (ka*c)


How to Find where to use these two equations ?

Answer 2

@electrokid @JFraser @shubhamsrg @Preetha @ghazi

Answer 3

\(K_a\)=ionization const.
\[HF\rightarrow H^++F^-\]
one -to one ratio

Answer 4

in Second case

CH3COOH ---> CH3COO- + H+

Answer 5

so, in 0.01M solution, what is the ionic concentration of \(H^+\)?

Answer 6

ionic concentration of H+ = Degree of Dissocaition

Answer 7

\[K_a={[H^+][F^-]\over[HF]}\]
but \([H^+]=[F^-]\)

Answer 8

u got my question ?

Answer 9

given that \(K_a=3.2\times10^{-4}\) and \([HF]=0.02M\), find \([H^+]\)

Answer 10

do you see the logic?

Answer 11

In Question I

HF --- > H+ + F-

we take Concentration of H+ = F- = c*alpha




In Question 2


CH3COOH ---> CH3COO- + H+

we take Concentration of CH3COO- = H+ = alpha

Answer 12

Why ?....Both the question are Same Type..

Answer 13

yes they all are

Answer 14

but The approach is Different ? Why ?

Answer 15

but information provided in the second one is different
\[K_a=\frac{\text{product of conc. of products}}{\text{product of conc. of reactants}}\]

Answer 16

ooh i see

Answer 17

yes..
you see, HF is a strong acid.. so \(K_a\) is for the forward reaction

Answer 18

on the other hand,
CH3COOH is a weak acid, so, \(K_a\) represents the bacward reaction
\[CH_3COOH\leftarrow CH3COO^-+H^+\]
so, what are your reactants and what your products/?

Answer 19

\[K_a=\frac{[CH_3COOH]}{[CH_3COO^-][H^+]}\]

Answer 20

That Make Sense/......THXxxxx

Answer 21

yep