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A monoprotic weak acid, HA , dissociates in water according to the reaction:
HA(aq) -----> H+(aq) + A−(aq)
The equilibrium concentrations of the reactants and products are:
[HA] = 0.200 M , [H+] = 4.00 x 10^− 4 M and [A −] = 4.00 x 10^− 4 M .
a. Calculate the value of pKa for the acid HA .

Answer :

Answer: The [tex]pKa[/tex] of the acid is 6.09

Explanation:

For the given chemical reaction:

[tex]HA(aq.)\rightleftharpoons H^+(aq.)+A^-(aq.)[/tex]

The expression of equilibrium constant [tex[(K_a)[/tex] for the above equation follows:

[tex]K_a=\frac{[H^+][A^-]}{[HA]}[/tex]

We are given:

[tex][HA]_{eq}=0.200M[/tex]

[tex][H^+]_{eq}=4.00\times 10^{-4}M[/tex]

[tex][A^-]_{eq}=4.00\times 10^{-4}M[/tex]

Putting values in above expression, we get:

[tex]K_a=\frac{(4.00\times 10^{-4})\times (4.00\times 10^{-4}}{0.200}\\\\K_a=8.0\times 10^{-7}0[/tex]

p-function is defined as the negative logarithm of any concentration.

[tex]pKa=-\log(K_a)[/tex]

So,

[tex]pKa=-\log(8.0\times 10^{-7})\\\\pKa=6.09[/tex]

Hence, the [tex]pKa[/tex] of the acid is 6.09

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