Here are some practice problems for writing K_{sp} expressions. Write the chemical equation showing how the substance dissociates and write the K_{sp} expression:
PART 1:
1) AlPO_{4}
2) BaSO_{4}
3) CdS
4) Cu_{3}(PO_{4})_{2}
5) CuSCN
6) Hg_{2}Br_{2}
7) AgCN
8) Zn_{3}(AsO_{4})_{2}
9) Mn(IO_{3})_{2}
10) PbBr_{2}
11) SrCO_{3}
12) Bi_{2}S_{3}
ANSWERS
Dissociation Equation K_{sp} expression AlPO_{4} <===> Al^{3+} (aq) + PO_{4}^{3}¯ (aq) K_{sp} = [Al^{3+}] [PO_{4}^{3}¯] BaSO_{4} <===> Ba^{2+} (aq) + SO_{4}^{2}¯ (aq) K_{sp} = [Ba^{2+}] [SO_{4}^{2}¯] CdS <===> Cd^{2+} (aq) + S^{2}¯ (aq) K_{sp} = [Cd^{2+}] [S^{2}¯] Cu_{3}(PO_{4})_{2} <===> 3 Cu^{2+} (aq) + 2 PO_{4}^{3}¯ (aq) K_{sp} = [Cu^{2+}]^{3} [PO_{4}^{3}¯]^{2} CuSCN <===> Cu^{+} (aq) + SCN¯ (aq) K_{sp} = [Cu^{+}] [SCN¯] Hg_{2}Br_{2} <===> Hg_{2}^{2+} (aq) + 2 Br¯ (aq) K_{sp} = [Hg_{2}^{2+}] [Br¯]^{2} AgCN <===> Ag^{+} (aq) + CN¯ (aq) K_{sp} = [Ag^{+}] [CN¯] Zn_{3}(AsO_{4})_{2} <===> 3 Zn^{2+} (aq) + 2 AsO_{4}^{3}¯ (aq) K_{sp} = [Zn^{2+}]^{3} [AsO_{4}^{3}¯]^{2} Mn(IO_{3})_{2} <===> Mn^{2+} (aq) + 2 IO_{3}¯ (aq) K_{sp} = [Mn^{2+}] [IO_{3}¯]^{2} PbBr_{2} <===> Pb^{2+} (aq) + 2 Br¯ (aq) K_{sp} = [Pb^{2+}] [Br¯]^{2} SrCO_{3} <===> Sr^{2+} (aq) + CO_{3}^{2}¯ (aq) K_{sp} = [Sr^{2+}] [CO_{3}^{2}¯] Bi_{2}S_{3} <===> 2 Bi^{3+} (aq) + 3 S^{2}¯ (aq) K_{sp} = [Bi^{3+}]^{2} [S^{2}¯]^{3}
Calculate the molar solubility (in mol/L) of a saturated solution of the substance.Silver chloride, AgCl, has a K_{sp} = 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 K_{sp} expression is as before:
K_{sp} = [Ag^{+}] [Cl¯]
This is the equation we must solve. First we put in the K_{sp} 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 x^{2} 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} | K_{sp} = 5.45 x 10¯^{27} |
Ag_{2}CrO_{4} | K_{sp} = 1.12 x 10¯^{12} |
Tin(II) hydroxide
Here is the equation for dissociation:
Sn(OH)_{2} <:===> Sn^{2+} + 2 OH¯
and here is the K_{sp} expression:
K_{sp} = [Sn^{2+}] [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 Sn^{2+} and OH¯ is one-to-two. That means that however much Sn^{2+} dissolves, we get DOUBLE that amount of OH¯. This is important, so go slow and think it through.
One Sn^{2+} makes two OH¯. That means that if 'x' Sn^{2+} 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 4x^{3} = 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:
Ag_{2}CrO_{4} <===> 2 Ag^{+} + CrO_{4}^{2}¯
K_{sp} = [Ag^{+}]^{2} [CrO_{4}^{2}¯] = 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:
4x^{3} = 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:
Bi_{2}S_{3} | K_{sp} = 1.82 x 10¯^{99} |
Cu_{3}(PO_{4})_{2} | K_{sp} = 1.93 x 10¯^{37} |
Bismuth sulfide
Bi_{2}S_{3} <===> 2 Bi^{3+} + 3 S^{2}¯
K_{sp} = [Bi^{3+}]^{2} [S^{2}¯]^{3}
1.82 x 10¯^{99} = (2x)^{2} (3x)^{3}
108x^{5} = 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
Cu_{3}(PO_{4})_{2} <===> 3 Cu^{2+} + 2 PO_{4}^{3}¯
K_{sp} = [Cu^{2+}]^{3} [PO_{4}^{3}¯]^{2}
1.93 x 10¯^{37} = (3x)^{3} (2x)^{2}
108x^{5} = 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
Na2SO4.
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 = [Pb^{2+}] [SO4^{2-}] = 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 Na2SO_{4}because of the common ion effect. |