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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 .

User Sal Rahman
by
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1 Answer

5 votes

Answer: The
pKa of the acid is 6.09

Step-by-step explanation:

For the given chemical reaction:


HA(aq.)\rightleftharpoons H^+(aq.)+A^-(aq.)

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


K_a=([H^+][A^-])/([HA])

We are given:


[HA]_(eq)=0.200M


[H^+]_(eq)=4.00* 10^(-4)M


[A^-]_(eq)=4.00* 10^(-4)M

Putting values in above expression, we get:


K_a=((4.00* 10^(-4))* (4.00* 10^(-4))/(0.200)\\\\K_a=8.0* 10^(-7)0

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


pKa=-\log(K_a)

So,


pKa=-\log(8.0* 10^(-7))\\\\pKa=6.09

Hence, the
pKa of the acid is 6.09

User Yawa Yawa
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