Question #6303e

1 Answer
Apr 13, 2017

Answer:

At 25 °C, 229 mg of #"Mg(OH)"_2# will remain undissolved.

Explanation:

This appears to be a solubility product problem.

We can set up an ICE table.

#color(white)(mmmmmm)"Mg(OH)"_2"(s)" ⇌ "Mg"^"2+""(aq)" + "2OH"^"-""(aq)"#
#"I/mol·L"^"-1": "color(white)(mmmmmmmmmmm)0color(white)(mmmmm)0#
#"C/mol·L"^"-1": color(white)(mmmmmmmmmm)"+"xcolor(white)(mmmm)"+2"x#
#"E/mol·L"^"-1": color(white)(mmmmmmmmmmll)xcolor(white)(mmmmm)2x#

At 25 °C,

#K_text(sp) = ["Mg"^"2+"]["OH"^"-"]^2 = 5.61×10^"-12"#

#x(2x)^2 = 4x^3 = 5.61×10^"-12"#

#x^3 = (5.61×10^"-12")/4 = 1.40 × 10^"-12"#

#x= 1.12 × 10^"-4"#

∴ The solubility of #"Mg(OH)"_2# at 25 °C is #1.12 × 10^"-4" color(white)(l)"mol/L"#.

#"Solubility" = (1.12 × 10^"-4" color(red)(cancel(color(black)("mol"))))/"1 L" × "58 320 mg"/(1 color(red)(cancel(color(black)("mol")))) = "6.53 mg/L"#

The solubility in 500 mL is 2.56 mg.

If you add 232 mg of #"Mg(OH)"_2# to 500 mL of water, the amount remaining undissolved will be

#"232 mg - 2.56 mg = 229 mg"#