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
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
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|
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]
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|
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.
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
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
Let molar solubility of PbSO4 be 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
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.