Question

Determine the Kanfor a .035 M weak (monoprotic) acid with a ph of 3.5. Show work too please?

Answer 1

weak monoprotic acid HA only weakly dissociates that means at equilibrium so of it is in the form HA and some of it is H+ and A-

HA dissociates into H+ and A-

HA <=======> H+ + A-

Ka is acid dissociation constant

lets do an ice chart to establish how much of each species is present at equilibrium

X amount of HA dissociates and X amount of H+ and X amount of A- is produced.

------ HA <======> H+ + A-
initial -- .060 ---------- 0 ---- 0
change - -X ---------- +X ---- +X
final ---- .035 - X ---- X ----- X

Ka = Products^coefficient/Reactants^coefficie...

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

let's plug in our values at equilibrium into the equation

Ka = [X][X]/[.035-X]

Since .035 is much >>>>> bigger than X we can ignore X in the denominator

Ka = [X]^2/.035

[X] = H+ concentration

the pH can be used to calculate the H+ con

the pH can be used to calculate the H+ concentration

pH = -log [H+]

[H+] = 10^-3.5 = X =3.625*10^-6

Plug in these values and you will get Ka. Hope it makes sense

ka= 3.625*10^-6 *3.625*10^-6 /0.035   = 375.46*10^-12   = 3.75*10^-10