Question #cbccf

1 Answer
Apr 15, 2017

Answer:

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"#