Fundamentals of Analytical Chemistry: 8th ed.
Chapter 11
Chapter 11 11-1
The overall dissociation constant for H2S is [ H 3O ]2 [S2- ] K1 K 2 [ H 2S]
In a solution saturated with the gas, [H2S] is constant, therefore [S2- ]
11-2
[ H 2S] K1 K 2 [ H 3 O ]2
or
[S2- ]
K [ H 3 O ]2
The simplifications in equilibrium calculations involve assuming that the concentrations of one or more species can be approximated as 0.00 M. Adding or subtracting a concentration that can be approximated as 0.00 leads to a meaningful result. In contrast, multiplying or dividing by 0.00 in the equilibrium constant expression causes the constant to become equal to zero or infinity. Thus, the expression is meaningless.
11-3
A charge-balance equation is derived by relating the concentration of cations and anions such that no. mol/L positive charge = no. mol/L negative charge For a doubly charged ion, such as Ba2+, the concentration electrons for each mole is twice the molar concentration of the Ba2+. That is, mol/L positive charge = 2[Ba2+] Thus, the molar concentration of all multiply charged species is always multiplied by the charge in a charge-balance equation.
11-4
(a)
0.20
=
[H3AsO4] + [H2AsO4-] + [HAsO42-] + [AsO43-]
(b)
0.10
=
[H3AsO4] + [H2AsO4-] + [HAsO42-] + [AsO43-]
2(0.10) =
[Na+] =
0.20
Fundamentals of Analytical Chemistry: 8th ed. (c)
11-6
0.0500 + 0.100
=
[ClO-] + [HClO]
0.100
=
[Na+]
[F-] + [HF]
=
0.25 + 2[Ca2+]
[Na+]
=
0.25
(e)
0.100 =
[Na+] =
(f)
[Ba2+] =
[C2O42-] + [HC2O4-] + [H2C2O4]
(g)
[Ca2+] =
½([F-] + [HF])
(a)
[H3O+] =
[OH-] + [H2AsO4-] + 2[HAsO42-] + 3[AsO43-]
(b)
[Na+] + [H3O+]
=
[OH-] + [H2AsO4-] + 2[HAsO42-] + 3[AsO43-]
(c)
[Na+] + [H3O+]
=
[OH-] + [ClO-]
(d)
[Na+] + [H3O+]+ 2[Ca2+]
=
[F-] + [OH-]
(e)
2[Zn2+] + [Na+] + [H3O+]
=
[OH-] + 2[Zn(OH)42-]
(f)
2[Ba2+] + [H3O+]
=
[OH-] + 2[C2O42-] + [HC2O4-]
(g)
2[Ca2+] + [H3O+]
=
[OH-] + [F-]
(d)
11-5
Chapter 11
[OH-] + 2[Zn(OH)42-]
Step 1 Ag2C2O4 2Ag+ + C2O42H2C2O4 + H2O H3O+ + HC2O4HC2O4- + H2O H3O+ + C2O42Step 2 S = solubility = [Ag+]/2 Step 3 [Ag+]2[C2O42-] = Ksp = 3.510-11
(1)
[ H 3O ][ HC 2 O -4 ] K1 5.60 10 2 H 2C 2O 4
(2)
[ H 3O ][C 2 O 24- ] K 2 5.42 10 5 [ HC 2 O 4 ]
(3)
Step 4 [Ag+] = 2([C2O42-] + [HC2O4-] + [H2C2O4])
(4)
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 11
[H3O+] = 1.0 10-6
(5)
Step 5 No charge-balance equation because a buffer of unknown composition is present. Step 6 Unknowns: [Ag+], [C2O42-], [HC2O4-], [H2C2O4] Equations: (1), (2), (3) and (4) Step 7 No approximations needed. Step 8 Substituting (5) into (3) and rearranging gives
1.0 10 6 C 2 O 24[HC2O4 ] = 5.42 10 5 -
= 0.01845[C2O42-]
Substituting this relationship and (5) into (2) and rearranging gives [H2C2O4] =
1.0 10 6 0.01845 C 2 O 245.60 10 2
= 3.2910-7[C2O42-]
Substituting these two relationships into (4) gives [Ag+] = 2[C2O42-] + 2(0.01845[C2O42-]) + 2(3.2910-7[C2O42-]) [Ag+] = 2.037[C2O42-]
or
[C2O42-] = 0.4909[Ag+] Substituting this relationship into (1) and rearranging gives [Ag+] = (3.510-11/0.4909)1/3 = 4.1510-4 M Finally S = 4.1510-4/2 = 2.110-4 Substituting other values for [H3O+] yields the following solubility data:
Fundamentals of Analytical Chemistry: 8th ed.
11-7
Chapter 11
[H3O+], M
Solubility, M
(a)
1.010-6
2.110-4
(b)
1.010-7
2.110-4
(c)
1.010-9
2.110-4
(d)
1.010-11
2.110-4
Proceeding as in Problem 11-6, we write BaSO4 Ba2+ + SO42-
Ksp = 1.110-10
HSO4- + H2O H3O+ + SO42-
K2 = 1.0210-2
S = [Ba2+] [Ba2+][SO42-] = 1.110-10
(1)
[ H 3O ][SO 24 ] 1.0210-2 [ HSO 4 ]
(2)
Mass balance requires that [Ba2+] = [SO42-] + [HSO4-]
(3)
[H3O+] = 2.5
(4)
Substituting (4) into (2) gives upon rearranging [HSO4-] = (2.5)[SO42-]/(1.0210-2) = 245.1[SO42-]
(5)
Substituting (5) into (3) gives [Ba2+] = [SO42-] + 245.1[SO42-] = 246.1[SO42-] Substituting (6) into (1) gives 246.1[SO42-]2 = 1.110-10 [SO42-] = 6.710-7 M From (6)
[Ba2+] = 1.610-4 M = S
(6)
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 11
Proceeding in a similar manner for the other [H3O+] gives
11-8
[H3O+]
S, mol/L
(a)
2.5
1.6510-4
(b)
1.5
1.2810-4
(c)
0.060
2.7510-5
(d)
0.200
4.7610-5
The following derivation applies to this and the following two problems. MS(s) M2+ + S2H2S + H2O H3O+ + HS-
K1 = 9.610-8
HS- + H2O- H3O+ + S2-
K2 = 1.310-14
H2S + 2H2O 2H3O+ + S2-
K1K2 = 9.610-8 1.310-14 = 1.2510-21
S = solubility = [M2+] = [S2-] + [HS-] + [H2S] [M2+][S2-] = Ksp
(1)
[ H 3O ][ HS ] K2 = 1.310-14 H 2S
(2)
[ H 3O ]2 [S 2 ] K1K2 = 1.2510-21 H 2S
(3)
From mass-balance consideration [M2+] = [S2-] + [HS-] + [H2S]
(4)
Substituting (2) and (3) into (4) gives [M2+] = [S2-] +
[ H O ] [ H 3 O ]2 [ H 3O ][S2 ] [ H 3O ]2 [S2 ] [S2 ]1 3 K2 K1 K 2 K2 K1 K 2
Substituting (1) into (5)
(5)
Fundamentals of Analytical Chemistry: 8th ed.
(a)
[M2+] =
K sp [ H 3O ] [ H 3O ]2 1 [ M 2 ] K2 K1 K 2
[M2+] =
[ H O ] [ H 3 O ]2 K sp 1 3 K K K 2 1 2
[M2+] =
[ H 3O ] [ H 3O ]2 K sp 1 14 1.25 10 21 1.3 10
Substituting Ksp = 810-37 and [H3O+] = 2.010-1 into (6) gives [M2+] = solubility =
0.202 0.20 8 10 37 1 14 1.25 10 21 1.3 10
= 5.1 10 9 M (b)
Substituting Ksp = 810-37 and [H3O+] = 2.010-4 into (6) gives solubility = 5.1 10 12 M
11-9
(a)
Substituting Ksp = 110-27 and [H3O+] = 2.010-1 into (6) gives solubility = 2 10 4 M
(b)
Substituting Ksp = 110-27 and [H3O+] = 2.010-4 into (6) gives solubility = 210-7 M
11-10 (a)
Substituting Ksp = 310-14 and [H3O+] = 2.010-5 into (6) gives solubility = 0.1 M
(b)
Substituting Ksp = 310-14 and [H3O+] = 2.010-7 into (6) gives solubility = 110-3 M
Chapter 11
(6)
Fundamentals of Analytical Chemistry: 8th ed. PbCO3 Pb2+ + CO32-
11-11
Chapter 11 Ksp = [Pb2+][CO32-] = 7.410-14
H2CO3 + H2O H3O+ + HCO3- K1 =
[ H 3O ][ HCO -3 ] 4.4510-7 [ H 2 CO 3 ]
HCO3- + H2O H3O+ + CO32-
[ H 3O ][CO 32- ] 4.6910-11 [ HCO 3 ]
K2 =
[Pb2+] = [CO32-] + [HCO3-] + [H2CO3] [H3O+] = 110-7 Proceeding as in problem 11-8 [M2+] =
2+
[Pb ] =
[ H O ] [ H 3 O ]2 K sp 1 3 K2 K1 K 2
7.4 10
14
7 1.0 10 7 2 1 1.0 10 4.69 10 11 4.45 10 7 4.69 10 11
solubility = [Pb2+] = 1.410-5 M 11-12 Ag2SO3 2Ag+ + SO32-
Ksp = [Ag+]2[SO32-] = 1.510-14
H2SO3 + H2O H3O+ + HSO3- K1 =
-
+
HSO3 + H2O H3O + SO3
2-
[ H 3O ][ HSO -3 ] 1.2310-2 [ H 2SO 3 ]
[ H 3O ][SO 32- ] 6.610-8 K2 = [ HSO3 ]
H2SO3 + 2H2O 2H3O+ + SO32-
K1K2 =
[ H 3O ]2 [SO 32- ] 8.110-10 [ H 2SO 3 ]
(1) (2)
(3)
(4)
The mass balances are ½ [Ag+] = [SO32-] + [HSO3-] + [H2SO3]
(5)
[H3O+] = 110-8
(6)
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 11
Substituting (2), (4) and (6) into (5) gives +
2-
½ [Ag ] = [SO3 ] +
1 10 [SO 8
6.6 10
23
]
-8
1 10 [SO 8 2
+
8.1 10
2 1 10 8 1 10 8 ½ [Ag ] = [SO3 ] 1 6.6 10 -8 8.1 10 10
+
2-
Substituting (1) into this expression gives ½ [Ag+] =
8 8 2 K sp 1 1 10 1 10 2 -8 8.1 10 10 Ag 6.6 10
1 108 2 1 10 8 [Ag+]3 = 21.510-14 1 6.6 10 -8 8.1 10 10 [Ag+] = 4.210-5 Solubility = 2.110-5 M 11-13 [Cu2+][OH-]2 = 4.810-20
[Mn2+][OH-]2 = 210-13
(a)
Cu(OH)2 precipitates first
(b)
Cu2+ begins to precipitate when [OH-] =
(c)
4.8 10 20 / 0.050 = 9.810-10 M
Mn2+ begins to precipitate when [OH-] =
2 10 13 / 0.040 = 2.210-6
[Cu2+] = 4.810-20/(2.210-6)2 = 9.610-9 M 11-14 Ba(IO3)2 Ba2+ + 2IO3BaSO4 Ba2+ + SO42To initiate precipitation of Ba(IO3)2
Ksp = 1.5710-9 Ksp = 1.310-10
-10
23
]
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 11
[Ba2+] = 1.5710-9/(0.050)2 = 6.310-7 To initiate precipitation of BaSO4 [Ba2+] = 1.310-10/(0.040) = 3.210-9 M (a)
BaSO4 precipitates first
(b)
[Ba2+] = 3.210-9 M
(c)
When [Ba2+] = 6.310-7 M [SO42-] = 1.110-10/(6.310-7) = 1.710-4 M
11-15 (a)
[Ag+] = Ksp/[I-] = 8.310-17/(1.010-6) = 8.310-11 M
(b)
[Ag+] = Ksp/[SCN-] = 1.110-12/(0.070) = 1.610-11 M
(c)
[I-] when [Ag+] = 1.610-11 M [I-] = 8.310-17/(1.610-11) = 5.210-6 M [SCN-]/[I-] = 0.070/(5.210-6) = 1.3104
(d)
[I-] = 8.310-17/(1.010-3) = 8.310-14 M [SCN-] = 1.110-12/(1.010-3) = 1.110-9 M [SCN-]/[I-] = 1.110-9/(8.310-14) = 1.3104 Note that this ratio is independent of [Ag+] as long as some AgSCN(s) is present.
11-16 (a)
[Ba2+][SO42-] = 1.110-10
[Sr2+][SO42-] = 3.210-7
BaSO4 precipitation is complete when [SO42-] = 1.110-10/(1.010-6) = 1.110-4 M SrSO4 begins to precipitate when [SO42-] = 3.210-7/(0.050) = 6.410-6 M SrSO4 begins to precipitate before the Ba2+ concentration is reduced to 1.010-6 M.
Fundamentals of Analytical Chemistry: 8th ed. (b)
[Ba2+][SO42-] = 1.110-10
Chapter 11
[Ag+]2[SO42-] = 1.610-5
BaSO4 precipitation is complete when [SO42-] = 1.110-4 M Ag2SO4 begins to precipitate when [SO42-] = 1.610-5/(0.020)2 = 0.040 M Ag2SO4 does not precipitate before the Ba2+ concentration is reduced to 1.010-6 M. (c) Be2+ precipitates when [OH-] = (7.010-22/(0.020))1/2 = 1.910-10 M Hf4+ precipitates when [OH-] = (4.010-26/(0.010))1/4 = 1.410-6 M Be precipitation complete when [OH-] = (7.010-22/1.010-6)1/2 = 2.610-8 M Hf(OH)4 does not precipitate before the Be2+ concentration is reduced to 1.010-6 M. (d)
In3+ precipitates when [IO3-] = (3.310-11/(0.20))1/3 = 5.510-4 M Tl+ precipitates when [IO3-] = 3.110-6/(0.090) = 3.410-5 M Tl+ precipitation complete when [IO3-] = 3.110-6/1.010-6 = 3.1 M
In(IO3)3 begins to precipitate before the Tl+ concentration is reduced to 1.010-6 M. 11-17 AgBr Ag+ + BrAg+ + 2CN- Ag(CN)2-
5.010-13 = [Ag+][Br-] 1.31021 =
[Ag (CN) 2 ] [Ag ][CN ]2
(1) (2)
It is readily shown that the reaction CN- + H2O HCN + OHproceeds to such a small extent that it can be neglected in formulating a solution to this problem. That is, [HCN] << [CN-], and only the equilibria shown need to be taken into account. Solubility = [Br-]
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 11
Mass balance requires that [Br-] = [Ag+] + [Ag(CN)2-]
(3)
0.100 = [CN-] + 2[Ag(CN)2-]
(4)
We now have 4 equations and 4 unknowns. Based upon the large size of 2 let us assume that [CN-] << 2[Ag(CN)2-]
and
[Ag+] << [Ag(CN)2-]
(4) becomes
[Ag(CN)2-] = 0.100/2 = 0.0500
and (3) becomes
[Br-] = [Ag(CN)2-] = 0.0500
To check the assumptions, we calculate [Ag+] by substituting into (1) [Ag+] = 5.010-13/0.0500 110-11
( 110-11 << 0.0500)
To obtain [CN-] we substitute into (2) and rearrange [CN-] =
0.0500 1 10 1.3 10 21
11
= 2.010-6
( 210-6 << 0.100)
Thus, the two assumptions are valid and Solubility = [Br-] = 0.0500 M mass AgBr/200 mL
= 0.0500
mmol 0.1877 g AgBr 200 mL mL mmol AgBr
= 1.877 g 11-18 CuCl(s) Cu+ + ClCu+ + 2Cl- CuCl2-
Ksp = [Cu+][Cl-] = 1.910-7 2 =
[CuCl-2 ] = 7.9104 - 2 [Cu ][Cl ]
(1) (2)
It is convenient to multiply (1) by (2) to give
[CuCl -2 ] = 1.910-77.9104 = 1.510-2 [Cl ]
(3)
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 11
From a charge balance consideration, we can write (if we assume [H3O+] = [OH-]) [Cu+] + [Na+] = [Cl-] + [CuCl2-]
(4)
By rearranging (1) and (3) and substituting into (4) we obtain [Cl-] = [Na+] +
1.9 107 - 1.510-2[Cl-] [Cl ]
which rearranges to the quadratic 0 = 1.015[Cl-]2 - [Na+][Cl-] - 1.910-7
(5)
By using [Na+] = the NaCl analytical concentration, (5) can be solved to give the following [Cl-] (a) 2.0 M
(c) 0.020 M
(b) 0.20 M
(d) 0.0021 M
(e) 5.410-4 M
Note that the equilibrium [Cl-] concentration is larger than the NaCl analytical concentration for parts (d) and (e). The reason for this apparent anomaly is that the dissolution of CuCl to give Cu+ and Cl- contributes significantly to the equilibrium [Cl-] at the lower NaCl analytical concentrations. The solubility of CuCl can be obtained from the calculated [Cl-] and the expression
1.9 107 S = [Cu ] + [CuCl2 ] = + 1.510-2[Cl-] [Cl ] +
-
Solution of this equation for each of the [Cl-] gives (a) 0.030 M
(c) 3.110-4 M
(b) 3.010-3 M
(d) 1.210-4 M
(e) 3.610-4 M
Fundamentals of Analytical Chemistry: 8th ed. 11-19 (a) CaSO4(s) Ca2+ + SO42CaSO4(aq) Ca2+ + SO42-
Chapter 11
Ksp = [Ca2+][SO42-] = 2.610-5
(1)
[Ca 2 ][SO 24- ] = 5.210-3 [CaSO 4 ( aq)]
(2)
Kd =
CaSO4(s) CaSO4(aq)
(3)
The mass balance gives [Ca2+] = [SO42-]
(4)
We have 3 equations and 3 unknowns ([Ca2+], [SO42-], and [CaSO4(aq)] To solve we divide (1) by (2) to give [CaSO4(aq)] = Ksp/Kd = (2.610-5)/(5.210-3) = 5.010-3 Note that this is the equilibrium constant expression for (3) and indicates that the concentration of un-ionized CaSO4 is always the same in a saturated solution of CaSO4. Substituting (4) into (1) gives [Ca2+] = (2.610-5)1/2 = 5.110-3 and since S = [CaSO4(aq)] + [Ca2+] we obtain S = 5.010-3 + 5.110-3 = 0.0101 M %CaSO4(aq) = (5.010-3/1.0110-2)100% = 49% (b) (Note: In the first printing of the text, the answer in the back of the book was in error.) Here [CaSO4(aq)] is again equal to 5.010-3 and the mass balance gives [SO42-] = 0.0100 + [Ca2+]
(5)
Substituting (1) into (5) and rearranging gives 0 = [SO42-]2 - 0.0100[SO42-] - Ksp which may be solved using the quadratic equation to give [SO42-] = 0.0121
and
[Ca2+] = 2.1410-3
Fundamentals of Analytical Chemistry: 8th ed. S = 5.010-3 + 2.1410-3 = 7.1410-3 %CaSO4(aq) = (5.010-3/7.1410-3)100% = 70%
Chapter 11