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

Answer : option c) heat is released during the process.explanation :

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.

What do you need with it’s not a question your just stating what it says