This homework is due at the start of class. If you wish to use your homework for

Question

This homework is due at the start of class. If you wish to use your homework for studying for the mid-semester exam, then make a photocopy of it before turning it in. The material covered in this homework assignment applies to processes that occur at the Earth's surface. Equilibrium constants that you may need are given below. Show all your work. A quantity of 0.5g of solid Na_2SO_4 (which is completely soluble and dissociates to Na^+ and SO_4^2) is added to one liter of a solution containing 1.5 times 10^-2 moles of dissolved Pb^2+. Demonstrate that anglesite (PbSO_4) precipitates from the solution and calculate the weight of PbSO_4 in grams that will precipitate per liter.

Explanation / Answer

We have 0.5 g of solid Na2SO4, which means we have:

0.5 g of Na2SO4 * (1mol / 142.04 g) = 0.00352 moles of Na2SO4

We have 1L solution with 0.015 moles of Pb+2 ions.

So concentrations will be, if we assume no change in volume:

[Pb+2] = 0.015 M

[SO4-2] = 0.00352 M

Now we get the ionic product:

IP = [Pb+2] [SO4-2] = 0.015 * 0.00352 = 5.28 x 10-5

A precipitate will form if ion product is greater than Ksp, and we have:

IP = 5.28 x 10-5 > Ksp = 1 x 10-7.8

So a precipitate will form, in the following quantity per liter:

First we make our ICE table:

To get equilibrium concentration, we use the definition of constant:

Ksp = [Pb+2] [SO4-2]

1 x 10-7.8 = (0.015 - x) (0.00352 - x)

Isolating for x:

x = [PbSO4] = 0.00351862 M = 0.00351862 mol / L

Passing moles to grams:

0.00351862 mol / L * (303.26 g/mol) = 1.067 g/L

Pb+2 (aq) + SO4-2 (aq) <-> PbSO4 (s) I 0.015 M 0.00352 M 0 M C -x -x +x E (0.015 - x) (0.00352 - x) x

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