If the ka of a monoprotic weak acid is 1.2 × 10-6, what is the ph of a 0.40 m solution of this acid?

First, we write out a balanced equation.
HA <--> H(+) + A(-)

Next, we create an ICE table

        HA     <-->      H+     +    A-
[]i     0.40M            0M           0M
Δ[]    -x                   +x            +x
[]f     0.40-x             x              x

Next, we write out the Ka expression.

Ka = [H+][A-]/[HA]

Ka = x*x/(0.40-x) 

However, because Ka is less than 10^-3, we can assume the amount of dissociation is negligible. Thus,

Assume 0.40-x ≈ 0.40

Therefore, 1.2x10^-6 = x^2/0.40

Then we solve for the [H+] concentration, or x



x=6.93x10^-4

Next, to find pH we do 

pH = -log[H+]

pH = -log[6.93x10^-4]

pH = 3.2

The density of water is 1 g/cm cubed and in order for the pebble to sink its density has to be greater than the density of water and in order for the twig to float its density has to be less than the density of water