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A certain weak acid, HA, has a Ka value of 9.9×10^-7
1. Calculate the percent dissociation of HA in a 0.10 M solution
2. Calculate the percent dissociation of HA in a 0.010 M solution
Ka in chemistry refers to an indicator of how much an acid breaks up. It is an equilibrium constant that is called acid dissociation/ionization constant. This constant provides information on the strength of acid. The greater it is the Ka number, the more powerful the acid. It’s based on the notion of strong acids being more likely to completely dissociate, resulting in high dissociation values for Ka. Strong acids break up completely while weak acids only dissociate only a small amount.
HA + H2O = H3O + A
For a 0.10M solution:
Where x is the concentration change:
Ka = [H3O] [A]/[HA]
9.9×10^-7 = x^2 / (0.10 – x)
x = 3.14 * 10.-4 mol/L
percent dissociation = x / 0.1 * 100 = 3.14 * 10^-1 percent
For a 0.010M solution:
Ka = [H3O] [A]/[HA]
9.9×10^-7 = x^2 / (0.010 – x)
x = 1.0 *10-5 mol/L
percent dissociation = x / 0.1 * 100 = 1.0 * 10^-2 percent
We all know that weak acids will cause dissociation.
Ha –>H+ + A-
The Ka = [H+][A] / [HA]
Initial conc. HA = 0.1
let at equilbirium
HA dissociated = x, so the conc will now be 0.1-x
[H+] =[A-] =x
So Ka = x 2 / 0.1-x
We can ignore x in deonimator because the acid is weak.
So
7.0