A 190 mL sample of 0.293 M NaCH3CO2(aq) is diluted to 750 mL. What is the concentration of the acetic acid at equilibrium? Ka =1.8×10−5.

Answer 1

NaC2H3O2 = NaAc
acetic acid = HC2H3O2 = HAc

NaAc diluted = 0.293 x (190/750) = approx 0.074 but you need to confirm this and all other calculations that follow. I estimate and round.
.......Ac^- + HOH ==> HAc + OH^-
I....0.074............0......0
C.......-x............x......x
E...0.074-x...........x......x

Kb for Ac^- = (Kw/Ka for HAc) =
(x)(x)/(0.074-x)
Solve for x = (HAc) = ?
I