If the 𝐾a of a monoprotic weak acid is 1.0×10−6, what is the pH of a 0.44 M solution of this acid?

Answer 1

HA ---> H+ + A-

I 0.44 0 0

C -x +x +x

E 0.44-x x x

Ka= x²/.44-x

because Ka is so small we are going to assume that 0.44-x is basically the same as 0.44 and cross out that x so we have

Ka= x²/0.44

1E-6(.44)=x^{^2}

x=6.6E-4

x is equal to our H+ concentration at equilibrium (according to our ICE table) therefore the pH is equal to -log(H+)

-log(6.6E-4)= 3.178