Question

Question #cbccf

Answer 1

The concentration of `"Pb"^"2+"` at equilibrium is `9.3 × 10^"-5"color(white)(l) "mol/L"`.

Explanation:

We can solve this by setting up an ICE table.

`color(white)(mmmmmm)"Pb(s)" + "2Cr"^"3+""(aq)" ⇋ "Pb"^"2+""(aq)" + "2Cr"^"2+""(aq)"`
`"I/mol·L"^"-1": color(white)(mmmmmm)0.100color(white)(mmmmm)0color(white)(mmmmml)0`
`"C/mol·L"^"-1": color(white)(mmmmmml)"-2"xcolor(white)(mmmmll)"+"xcolor(white)(mmmm)"+"2x`
`"E/mol·L"^"-1": color(white)(mmmmm)"0.100-2"xcolor(white)(mmmm)xcolor(white)(mmmmm)2x`

`K_text(c) = (["Pb"^"2+"]["Cr"^"2+"]^2)/(["Cr"^"3+"]^2) = (x(2x)^2)/(0.100-2x)^2 =(4x^3)/(0.100-2x)^2 = 3.2 × 10^"-10"`

Check for negligibility:

`0.100/(3.2 × 10^"-10") = 3.01 × 10^8 > 400`

∴ `2x` is negligible compared to 0.100.

Then

`K_text(c) = (4x^3)/0.100^2 = 3.2 × 10^"-10"`

`4x^3 = 0.100^2 × 3.2 × 10^"-10" = 3.2 × 10^"-12"`

`x^3 = 8.0 × 10^"-13"`

`x = 9.3 × 10^"-5"`

`["Pb"^"2+"] = x color(white)(l)"mol/L" = 9.3 × 10^"-5"color(white)(l) "mol/L"`