Question #31ba5

1 Answer
Feb 14, 2018

Answer:

Here's what I got.

Explanation:

Lead(II) bromide is only sparingly soluble in water, so when you place a sample of lead(II) bromide in water, you should expect an equilibrium to be established in aqueous solution between the undissolved solid and the dissolved ions.

#"PbBr"_ (2(s)) rightleftharpoons "Pb"_ ((aq))^(2+) + 2"Br"_ ((aq))^(-)#

Now, the solubility product constant, #K_(sp)#, of lead(II) bromide is calculated by using the equilibrium concentrations of the dissolved ions in a saturated solution of lead(II) bromide.

#K_(sp) = ["Pb"^(2+)] * ["Br"^(-)]^2#

Notice that each mole of lead(II) bromide that dissolves produces #1# mole of lead(II) cations and #2# moles of bromide anions.

This means that if you take #s# #"M"# to be the molar solubility of the salt, i.e. the number of moles of lead(II) bromide that can be dissolved in #"1 L"# of solution to produce a saturated solution, you can say that you have

#["Pb"^(2+)] = s quad "M"#

#["Br"^(-)] = 2 * s quad "M"#

This means that the solubility product constant will be equal to

#K_(sp) = s * (2s)^2#

#K_(sp) = 4s^3#

Rearrange the equation to solve for #s#

#s = root(3)(K_(sp)/4)#

Plug in the value you have for the solubility product constant to get

#s = root(3)((4 * 10^(-6))/4) = 1 * 10^(-2)#

You can thus say that the molar solubility of lead(II) bromide is equal to

#s = color(darkgreen)(ul(color(black)(1 * 10^(-2) quad "M")))#

The answer is rounded to one significant figure.

if you want, you can calculate the solubility of the salt in grams per liter by using the molar mass of lead(II) bromide.

#1 * 10^(-2) color(red)(cancel(color(black)("moles PbBr"_2))) * "367.1 g"/(1color(red)(cancel(color(black)("mole PbBr"_2)))) = color(darkgreen)(ul(color(black)("4 g L"^(-1))))#

Once again, the answer is rounded to one significant figure.