Question

Although the formula of this new chemical is a trade secret, it can be revealed that the formula for Herbigon is X-acetate (XCH3COO, where "X" represents the top-secret cation of the salt). It is this cation that kills weeds. Since it is critical to have Herbigon dissolved (it won't kill weeds as a suspension), you are working on adjusting the pH so Herbigon will be soluble at the concentration needed to kill weeds. What pH must the solution have to yield a solution in which the concentration of X+ is 3.50×10⁻³M ? The pKa of acetic acid is 4.76.

Answer 1

I think you missed to mention the Ksp of salt which I am assuming as 10^-5 ( you can put your value and do the calculation accrodingly)

Ksp = [X+][CH3COO-]
10^-5 = (3.50 x 10^-3)[CH3COO-]
[CH3COO-] =2.85 x 10^-3 M = 0.00285 M

The solution of weak acid with its conjugate salt will form buffer. The CH3COOH will dissociate as ( concentration of acetic acid = 1M)

Molarity . . . . . .CH3COOH + H2O <==> H3O+ + CH3COO-
initial . . . . . . . . . . .1 . . . . . . .. . ... . . . . . . .0 . . . . . .0.00285
change . . . . . . . . .-x . . . . . . . . . . . . . . . . . x . . . . . . . . .x
at equilibrium . . . .1-x . . . . . . . . . . . . . . . . ..x . . . .0.00285+x

pKa = 4.76;

We know that pka = -logKa

So Ka = 1.75 x 10^-5

Ka = 1.75 x 10^-5 = ([H3O+][CH3COO-] / [CH3COOH]) = (x)(0.00285x) / (1-x)

we can ignore x in denominator as Ka is small.
1.75 x 10^-5 = (x)(0.00285+x)
x^2 + 0.00285x - 1.75 x 10^-5 = 0

solving for x

So x = 0.0029 = [H3O+]
pH = -log [H3O+] = -log (0.0029) = 2.53