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Ksp Problems – Chemistry
Name: _______________________________________
- The value of Ksp of AgCl is 1.8 x 10
- . What would be the molar concentration of Ag + and Cl - in pure water placed in contact with solid AgCl(s)? Ksp = [Ag
] [ Cl
] 1.8 x 10-10 = [x] [x] x = 0.000013416 M or 1.3416 x 10-5^ M 2 ) The molar solubility of PbCl 2 in 0.10 M NaCl is 1.7 x 10-3^ moles in a liter (that is 1.7 x 10-3^ moles of PbCl 2 will dissolve in 1 liter of the solution). What is the Ksp of PbCl 2? PbI 2 ↔ Pb
Ksp = [Pb+] [ Cl-]^2 Ksp = [1.7 x 10
] [3.4 x 10
] 2 Ksp = 1.965 x 10 - 3 ) What are the molar concentrations of ions in solution when solid PbI 2 is in contact with pure water? (Ksp for PbI 2 = 7.9 x 10-9 ) PbI 2 ↔ Pb
Ksp = [Pb+] [ I-]^2 7.9 x 10
= [x] [ 2 x] 2 (a) 1.8 x 10
M (b) 1.3 x 10
M (c) 3.24 x 10 -20^ M (a) [Pb
] = 1.3 x 10
M & [I
] = 2.6 x 10
M (b) [Pb+2 ] = 1.6 x 10-3 M & [I- ] = 3.2 x 10-3 M (c) [Pb
] = [I
] = 2.0 x 10
M (a) Ksp = 1.8 x 10
(b) Ksp = 1.9 x 10- (c) Ksp = 1.7 x 10
3.95 x 10
M Pb
3.162 x 10
M I -** 4 ) What is the molar concentration of [Ag
] in AgCl solution in 0.10 M NaCl? (Ksp = 1.8 x 10-10 ) Remember that in this case the molar solubility of AgCl is equal to the [Ag+ ] as only the Ag
reflects the amount of AgCl that dissolved. Ksp = [Ag+] [ Cl-] 1.8 x 10
= [x] [x] x = 0.000013416 M or 1.3416 x 10-5 M
- Will CaSO 4 precipitate if 50 mL of 0.0010 M CaCl 2 is added to 50 mL of 0.010 M Na 2 SO 4? (Ksp for CaSO 4 = 2.4 x 10
) (a) Ag+ = 1.3 x 10
M (b) Ag+ = 0.10 M (c) Ag+ = [AgCl] = 1.8 x 10-9 M (a) no precipitation will occur (b) it is at equilibrium (c) precipitation will occur
Ksp = [Ag
][Cl
- ] = [x][1] = 1.8x - x = 1.8x - mol/L or 2.6x - g/L
Precipitation of Ionic Compounds
Will precipitate form if solutions mixed in 1:1 ratio? Ksp for AgCl = 1.8x
After mixing, [Ag
] = [Cl
] = 1.0x
M Now determine the ion product (IP) IP = [Ag+1][Cl-1] = [1.0x10-5][1.0x10-5] = 1.0x10- Compare IP to Ksp: If IP > Ksp precipitate forms If IP < Ksp no precipitate forms If IP = Ksp no precipitate forms but solution is saturated For this problem, IP < Ksp and no precipitate forms! Sample Problem # Will mixture precipitate if equal volumes of 3.0x
M Ba
& 2.0x
M CO 3
mixed? Ksp for BaCO 3 is 5.0x10- After mixing, [Ba+2] = 1.5x10-3M [CO 3 -2] = 1.0x10-3M IP = [Ba
][CO 3
] = [1.5x
][1.0x
] = 1.5x
Since IP > Ksp, precipitate forms! Will mixture precipitate if 20.0mL 3.0x
M Ba
& 30.0mL 2.0 x
M CO 3
mixed? After mixing, M 1 V 1 =M 2 V 2 (3.0x
)(20mL) = M 2 (50mL) [Ba
] = 1.2x
M M 1 V 1 =M 2 V 2 (2.0x10-4)(30mL) = M 2 (50mL) [CO 3 -2] = 1.2x10-4M IP = [Ba+2][CO 3 -2] = [1.2x10-4][1.2x10-4] = 1.4x10- Since IP > Ksp, precipitate forms! Sample Problem # Is 0.10M Mg(NO 3 ) 2 soluble at pH of 12? [Mg
] = 0.10M [OH
] = 0.01M Mg(OH)2(s) Mg
Ksp = [Mg
][OH
] 2 = 7.1x
IP = [Mg+2][OH-1]^2 = [0.10][0.01]^2 = 1.0x10- Since IP > Ksp, precipitate forms! Sample Problem # Is 0.10M Mg(NO 3 ) 2 soluble at pH of 8 [Mg
] = 0.10M [OH
] = 1.0x
M Mg(OH)2(s) Mg+2 + 2OH-1 Ksp = [Mg+2 ][OH-1 ]^2 = 7.1x10- IP = [Mg+2][OH-1]^2 = [0.10][1.0x10-6]^2 = 1.0x10- Since IP < Ksp, no precipitate forms! Sample Problem #
Determine optimum conditions for separating 0.10 Mg
& 0.10M Ca
The ions can be separated by adjusting pH since Ca(OH) 2 Ksp = 6.5x10-6^ more soluble Mg(OH) 2 Ksp = 7.1x
- less soluble Adjust pH to make saturated solution of Ca(OH) 2 but do not precipitate Ca(OH) 2 To make saturated solution of Ca(OH) 2 : Ksp = [Ca+2][OH-1]^2 = [0.10][OH-1]^2 = 6.5x10- [OH
] 2 = 6.5x
[OH
] = 8.1x
pOH = 2.09 pH =11. **Adjust pH at 11.91 Ca
does not precipitate but Mg
will ppt Sample Problem #** Determine optimum conditions to separate 0.10M Ni
& 0.10M Sr
by precipitating with Na 2 CO 3 NiCO 3 Ksp = 1.3x
more soluble SrCO 3 Ksp = 9.3x10-10^ less soluble Add CO 3 -2^ to make saturated solution of NiCO 3 but do not precipitate NiCO 3 To make saturated solution of NiCO 3 : Ksp = [Ni+2][CO 3 -2] = [0.10][CO 3 -2] = 1.3x10- [CO 3
] = 1.3x
M **Adjust Na 2 CO 3 concentration at 1.3x
M Ni
does not precipitate but Sr
will ppt**
