Question

The solubility of silver acetate in pure water at `25^o`C is `8.35 g/L` and `61.8 g/L` in an acid solution of `pH=3`. Calculate?

(i) `K_(sp)` of silver acetate.
(ii) dissociation constant of acetic acid.

Answer 1

Warning! Long Answer. (i) `K_text(sp) = 2.50 × 10^"-3"`; (ii) `K_text(a) = 1.82 × 10^"-5"`

Explanation:

This problem is a disguised Hess' Law problem using equilibrium constants.

(i) `K_text(sp)` for silver acetate

`s = (8.35 color(red)(cancel(color(black)("g"))))/"1 L" × ("1 mol")/(166.91 color(red)(cancel(color(black)("g")))) = 5.003 × 10^"-2" color(white)(l)"mol/L"`

`color(white)(mmmmmm)"AgOAc" ⇌ "Ag"^"+" + "AcO"^"-"`
`"E/mol·L"^"-1": color(white)(mmmmmmm)scolor(white)(mmm)s`

`K_text(sp) = ["Ag"^"+"]["AcO"^"-"] = s^2 = (5.003 × 10^"-2")^2 = 2.50 × 10^"-3"`

At pH 3

`s = (61.8 color(red)(cancel(color(black)("g"))))/"1 L" × "1 mol"/(166.91 color(red)(cancel(color(black)("g")))) = 3.702 × 10^"-1" color(white)(l)"mol/L"`

The equilibrium reaction is

`color(white)(mmmmmm)"AgOAc" + "H"_3"O"^"+" ⇌ "Ag"^"+" + "HOAc" + "H"_2"O"`
`"E/mol·L"^"-1": color(white)(mmmmll)1.0 ×10^"-3" color(white)(mll)scolor(white)(mmmll)s`

`K_text(eq) = (["Ag"^"+"]["HOAc"])/(["AcO"^"-"]) = s^2/(1.0 × 10^"-3") = (3.702 × 10^"-1")^2/(1.0 × 10^"-3")`

`= 1.371 × 10^2`

(ii) `K_text(a)` for acetic acid

We have two equations:

`bb(1.) color(white)(m)"AgOAc" ⇌ "Ag"^"+" + "AcO"^"-"; color(white)(mmmmmmmmll)K_1 = 2.503 × 10^"-3"`
`bb(2.) color(white)(m)"AgOAc" + "H"_3"O"^"+" ⇌ "Ag"^"+" + "HOAc" + "H"_2"O"; K_2 = 1.371 × 10^2`

From these, we must construct the target equation.

`"HOAc + H"_2"O" ⇌ "H"_3"O"^"+" + "AcO"^"-"`

The target equation has `"HOAc"` on the left, so we reverse Equation 2.

`bb(3.) color(white)(m)"Ag"^"+" + "HOAc" + "H"_2"O" ⇌ "AgOAc" + "H"_3"O"^"+"; K_3 = 1/K_2`

The target equation has `"AgOAc"` on the right, so we rewrite equation 1.

`bb(4.) color(white)(m)"AgOAc" ⇌ "Ag"^"+" + "AcO"^"-"; color(white)(mmmmmmmmll)K_4 = K_1`

Now, we add Equations 3. and 4.

`bb(3.) color(white)(m)color(red)(cancel(color(black)("Ag"^"+"))) + "HOAc" + "H"_2"O" ⇌ color(red)(cancel(color(black)("AgOAc"))) + "H"_3"O"^"+"; K_3 = 1/K_2`
`bb(4.) color(white)(m)ul(color(red)(cancel(color(black)("AgOAc"))) ⇌ color(red)(cancel(color(black)("Ag"^"+"))) + "AcO"^"-"color(white)(mmmmm)); color(white)(mmmll)ul(K_4 = K_1color(white)(m))`
`bb(5.) color(white)(m)"HOAc" + "H"_2"O" ⇌ "H"_3"O"^"+" + "AcO"^"-"; color(white)(mmmmmm)K_text(a) = K_3K_4 = K_1/K_2`

`K_text(a) = (2.50× 10^"-3")/(1.371 ×10^2) = 1.82 × 10^"-5"`