Calculate the pH and the concentrations of all species present (H3O+, F-, HF and OH-) in 0.05 M HF.
Let α be the dissociation of the weak acid
HF + H2O <---> H3O + +
F-
initial conc. c 0 0
change -cα +cα +cα
Equb. conc. c(1-α) cα cα
Dissociation constant , Ka = cα x cα / ( c(1-α)
= c α2 / (1-α)
In the case of weak acids α is very small so 1-α is taken as 1
So Ka = cα2
==> α = √ ( Ka / c )
Given Ka = 6.6x10-4
c = concentration = 0.05 M
Plug the values we get α = 0.1149
[H3O+] = cα = 0.05x0.1149 = 5.74x10-3 M
[ F-] = cα = 0.05x0.1149 = 5.74x10-3 M
pH = - log [H3O+]
= - log( 5.74x10-3)
= 2.24
[HF] = c(1-α) = 0.044 M
[OH-] = (1.0x10-14) / [H3O+] = 1.74x10-12 M
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