The Ka of a monoprotic weak acid is 7 84 10 3 What is thepercent ion...

Let a be the dissociation of the weak acid HA H A initial conc c 0 0 cha ....

The Ka of a monoprotic weak acid is 7.84 Ã— 10-3. What is the percent ionization...

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Let a be the dissociation of the weak acid
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â HA <---> H ⁺ + A^-

initial conc.Â Â Â Â Â Â Â Â Â Â Â cÂ Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â 0

changeÂ Â Â Â Â Â Â Â Â Â Â Â Â Â -caÂ Â Â Â Â Â Â Â Â Â Â +caÂ Â Â Â Â +ca

Equb. conc.Â Â Â Â Â Â Â Â c(1-a)Â Â Â Â Â Â Â Â caÂ Â Â Â Â ca

Dissociation constant , K_a = ca x ca / ( c(1-a)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = c a² / (1-a)

In the case of weak acids Î± is very small so 1-a is taken as 1

So K_a = ca²

==> a = âˆš ( K_a / c )

Given K_a = 7.84x10^-3

Â Â Â Â Â Â Â Â Â c = concentration = 0.149 M

Plug the values we get a = 0.229

âˆ´ % dissociation = 0.229 x 100 = 22.9

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