## Some uses of Ksp with problems in Chapter 16

(C) - Copyright, 2001 F.W. Boyle, Jr., Ph.D.

**Problem 16.6**

Complete the following table for concentrations in equilibrium
with PbBr_{2}.

| [Pb^{2+}] | [Br^{-}] |
---|

a) | | 0.007 |

b) | 0.4 | |

c) | | 1/2[Pb^{2+}] |

d) | 4[Br^{-}] | |

First consider what Ksp implies. It implies the [ ] of ions
dissolved in solution in equilibrium with some solid which
contains the ions.
In this case was have PbBr_{2}. We look up the Ksp in a
table and find Ksp = 6.6 x 10^{-6}

Next we must determine the dissolution (dissolving) reaction for
the compound.

PbBr_{2} ^{¬}_{®} Pb^{2+} + 2Br^{-}

Next we must write the Ksp expression.

Ksp = [Pb^{2+}][Br^{-}]^{2}

This expression is numerically equal to the value 6.6 x 10^{-
6}.
So:

Ksp = [Pb^{2+}][Br^{-}]^{2} = 6.6 x 10^{-6}

You should notice that there are two (2) unknowns in this
expression, [Pb^{2+}] and [Br^{-}]. You should
also notice that the table provides one concentration for each of
its four parts. So simply substitute in the expression and
solve.
a) [Br^{-}] = 0.007 substituting:

[Pb^{2+}](0.007)^{2} = 6.6 x 10^{-6}

And solving: [Pb^{2+}] = 0.1 __M__
You probably got 0.135 __M__ but forgot there was only one
significant figure in 0.007.

This is the [Pb^{2+}] that can be in solution with Br at 0.007
__M__ with PbBr_{2} solid.

b) [Pb^{2+}] = 0.4 __M__ substituting:

(0.4)[Br^{-}]^{2} = 6.6 x 10^{-6}

And solving: [Br^{-}] = 4 x 10^{-3} __M__
c) [Br^{-} = 1/2[Pb^{2+}]

This looks harder but really isn't!

Substituting:

[Pb^{2+}](1/2[Pb^{2+}])^{2} = 6.6 x 10^{-6}

Next square the second portion of the left side and:

[Pb^{2+}](1/4)[Pb^{2+}]^{2} = 6.6 x 10^{-6}

Then combine terms as you would in any mathematical
expression:

(1/4)[Pb^{2+}]^{3} = 6.6 x 10^{-6}

Multiply by 4 and take the cube root:
[Pb^{2+}] = 3 x 10^{-2} __M__ or 0.03
__M__

d) I will leave to you to solve!

**Problem 16.18:**

Calculate the solubility (in grams per liter) of MgF_{2}
in:

a) pure water

b) 0.015 __M__ Mg(NO_{3})_{2}

c) 0.0050 __M__ SbF_{3}

Part a) is a simple solubility. Write out the dissolution
reaction.

MgF_{2} ^{¬}_{®} Mg^{2+} + 2F^{-}

Next write the Ksp expression and go look up the Ksp.

Ksp = [Mg^{2+}][F^{-}] = 7 x 10^{-11}

Now when MgF_{2} dissolves it release 1 Mg and 2 F ions.
If we call the concentration of Mg, s then the concentration of F
is 2 s.
Substituting:

[s][2s] = 7 x 10^{-11}

Solving we find:

s = (Ksp/4)^{1/3}

This is the same as the XY_{2} form.
Plugging in Ksp and solving we find:

s = [Mg^{2+}] = 2.6 x 10^{-4} __M__

This is 0.02 g of MgF_{2}.
Part b) is a study of the common ion effect. Common ion means an
ion is dissolving into solution from two different sources. For
part b) the common ion is Mg. The NO_{3}^{-} has
absolutely nothing to do with solving this problem.

The Ksp is still 7 x 10^{/11} and the expression is
still:

Ksp = [Mg^{2+}][F^{-}] = 7 x 10^{-11}

However in this case we have to figure out the dissolution
reaction for Mg(NO_{3})_{2}.
Mg(NO_{3})_{2} ^{¬}_{®} Mg^{2+} + 2NO_{3}^{-}
So 1 Mg(NO_{3})_{2} releases 1 Mg^{2+}
and 2 F^{-}. Thus 0.015 __M__ releases 0.015
__M__ Mg^{2+}.

The MgF_{2} also releases some Mg and some F. How much?
We don't know so we call it **x**. Thus Mg is **0.015 + x**
and F is equal to **x**.

Back to our Ksp expression where we substitute the amount of
Mg^{2+}:

(0.015+x)(x) = 7 x 10^{-11}

But we can guess from part a that 0.015 is large compared to
**x** and we ignore the **x **added to 0.015. Now the equation is
simple and we solve for **x**.
And solve for x which equals [F^{-}]

[F^{-}] = (7 x 10^{-11}/0.015 )^{1/2}
= 6.8 x 10^{-5}

Now we must recall that each MgF_{2} releases 2 F^{-
} ions so the moles of MgF_{2} is half the number of
moles of F^{-}.
So:

MgF_{2} = 6.8 x 10^{-5}/2 = 3.4 x 10^{-5} moles

The problem wants to know the g so we must convert moles to
grams.
molar mass of MgF_{2} is 62 so multiply 62 x 3.4 x
10^{-5} which results in 2 x 10^{-3} g.

c) Try it yourself!