## SOLUBILITY PROBLEMS

Here are some practice problems for writing Ksp expressions. Write the chemical equation showing how the substance dissociates and write the Ksp expression:

### PART 1:

1) AlPO4
2) BaSO4
3) CdS
4) Cu3(PO4)2
5) CuSCN
6) Hg2Br2
7) AgCN
8) Zn3(AsO4)2
9) Mn(IO3)2
10) PbBr2
11) SrCO3
12) Bi2S3

 Dissociation Equation Ksp expression AlPO4 <===> Al3+ (aq) + PO43¯ (aq) Ksp = [Al3+] [PO43¯] BaSO4 <===> Ba2+ (aq) + SO42¯ (aq) Ksp = [Ba2+] [SO42¯] CdS <===> Cd2+ (aq) + S2¯ (aq) Ksp = [Cd2+] [S2¯] Cu3(PO4)2 <===> 3 Cu2+ (aq) + 2 PO43¯ (aq) Ksp = [Cu2+]3 [PO43¯]2 CuSCN <===> Cu+ (aq) + SCN¯ (aq) Ksp = [Cu+] [SCN¯] Hg2Br2 <===> Hg22+ (aq) + 2 Br¯ (aq) Ksp = [Hg22+] [Br¯]2 AgCN <===> Ag+ (aq) + CN¯ (aq) Ksp = [Ag+] [CN¯] Zn3(AsO4)2 <===> 3 Zn2+ (aq) + 2 AsO43¯ (aq) Ksp = [Zn2+]3 [AsO43¯]2 Mn(IO3)2 <===> Mn2+ (aq) + 2 IO3¯ (aq) Ksp = [Mn2+] [IO3¯]2 PbBr2 <===> Pb2+ (aq) + 2 Br¯ (aq) Ksp = [Pb2+] [Br¯]2 SrCO3 <===> Sr2+ (aq) + CO32¯ (aq) Ksp = [Sr2+] [CO32¯] Bi2S3 <===> 2 Bi3+ (aq) + 3 S2¯ (aq) Ksp = [Bi3+]2 [S2¯]3

### PART 2:

Calculate the molar solubility (in mol/L) of a saturated solution of the substance.
Silver chloride, AgCl, has a Ksp = 1.77 x 10¯10. Calculate its solubility in moles per liter.

The dissociation equation is as before:

AgCl (s) <===> Ag+ (aq) + Cl¯ (aq)

The Ksp expression is as before:

Ksp = [Ag+] [Cl¯]

This is the equation we must solve. First we put in the Ksp value:

1.77 x 10¯10 = [Ag+] [Cl¯]

Now, we have to reason out the values of the two guys on the right, First of all, I have no idea what the values are so I'll use unknowns. Like this:

I examine the chemical equation and see that there is a one-to-one ratio between Ag+ and Cl¯. I know this from the coefficients (both one) of the balanced equation.

That means that the concentrations of the two ions are EQUAL. I can use the same unknown to represent both. Like this:

[Ag+] = x = [Cl¯]

Substituting, we get:

1.77 x 10¯10 = (x) (x)

Now, we take the square root of both sides. I hope I'm not too insulting when I emphasize both sides. I have had lots of people take the square root of the x2 side, but not the other. After the square root, we get:

x = 1.33 x 10¯5 M

This is the answer because there is a one-to-one relationship between the Ag+ dissolved and the AgCl it came from. So, the molar solubility of AgCl is 1.33 x 10¯5 moles per liter.

Calculate the molar solubility (in mol/L) of a saturated solution of the substance.
However, there is additional explaining to do when compared to the AgCl example.

Here are the two substances:

 Sn(OH)2 Ksp = 5.45 x 10¯27 Ag2CrO4 Ksp = 1.12 x 10¯12

Tin(II) hydroxide

Here is the equation for dissociation:

Sn(OH)2 <:===> Sn2+ + 2 OH¯

and here is the Ksp expression:

Ksp = [Sn2+] [OH¯]2

So far, nothing out of the ordinary. However, that two in front of the hydroxide will come into play real soon.

The ratio between Sn2+ and OH¯ is one-to-two. That means that however much Sn2+ dissolves, we get DOUBLE that amount of OH¯. This is important, so go slow and think it through.

One Sn2+ makes two OH¯. That means that if 'x' Sn2+ dissolves, then '2x' of the OH¯ had to have dissolved.

We now write an equation:

5.45 x 10¯27 = (x) (2x)2

So we have 4x3 = 5.45 x 10¯27

and solving that, we get:

x = 1.11 x 10¯9 M = [Sn]

2 x = 2.22 x 10¯9 M = [OH]

Silver chromate

Here is the usual info:

Ag2CrO4 <===> 2 Ag+ + CrO42¯
Ksp = [Ag+]2 [CrO42¯] = 1.12 x 10¯12

In this problem the cofficient of 2 is on the first ion, not the second. No problem!

1.12 x 10¯12 = (2x)2 (x)

We now have:

4x3 = 1.12 x 10¯12

which is solved to give the answer:

x = 6.54 x 10¯5 M.

Calculate the molar solubility (in mol/L) of a saturated solution of the substance.

The two example substances are:

 Bi2S3 Ksp = 1.82 x 10¯99 Cu3(PO4)2 Ksp = 1.93 x 10¯37

Bismuth sulfide

Bi2S3 <===> 2 Bi3+ + 3 S2¯

Ksp = [Bi3+]2 [S2¯]3

1.82 x 10¯99 = (2x)2 (3x)3

108x5 = 1.82 x 10¯99

x = 7.00 x 10¯21 M

Note how 'x' still is the moles of the substance that dissolved.

Copper(II) phosphate

Cu3(PO4)2 <===> 3 Cu2+ + 2 PO43¯

Ksp = [Cu2+]3 [PO43¯]2

1.93 x 10¯37 = (3x)3 (2x)2

108x5 = 1.93 x 10¯37

x = 1.78 x 10¯8 M

### PART 3:

Calculating The Effect of a Common Ion on Solubility

 Exercise 6: Using the Ksp value from Exercise 1 calculate the solubility of PbSO4 in 0.100 M Na2SO4. Let molar solubility of PbSO4 be x. Ksp = [Pb2+] [SO42-] = = 1.96´ 10-8 PbSO4(s) + aq  [PbSO4(dissolved)] ®                   Pb2+(aq) +       SO42-(aq)                                                            x                                        x                   0.100 +x Since x is very small compared to 0.100, 0.100 +x is very nearly equal to 0.100.     Ksp = [Pb2+] [SO42-] = x ´ 0.100 = 1.96´ 10-8 \ x =                        = 1.96´ 10-7 M Molar solubility of PbSO4 in 0.100 M Na2SO4 = 1.96´ 10-7 Molar solubility of PbSO4 in pure water = 1.40´10-4 Note the solubility is very much reduced in 0.100 M Na2SO4because of the common ion effect.

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