Transcript Double Indicator Titration & Solubility Product (Ksp)

p.01
Acid-Base Eqm (7):
Double Indicator Titration
& Solubility Product (Ksp)
Double Indicator Titration
[Phenolphthalein & Methyl Orange]
For mixtures containing TWO BASES.
(e.g. NaOH / Na2CO3 / NaHCO3)
decreasing Kb
When titrate against standard HCl(aq):
NaOH + HCl  NaCl + H2O
Na2CO3 + HCl  NaHCO3 + NaCl
NaHCO3 + HCl  NaCl + CO2 + H2O
C. Y. Yeung (CHW, 2009)
All CO32- converted to HCO3-.
p.02
11.2 cm3 0.1M HCl: Phenolphthalein changes colour.
28.8 cm3 0.1M HCl: Methyl Orange changes colour.
All HCO3- converted to CO2 and H2O.
E.g.
25cm3 mixture containing
to be neutralized first
NaHCO3 & Na2CO3
11.2 cm3 0.1M HCl
NaHCO3
no. of mol of Na2CO3
= 1.1210-3 mol
28.8 cm3 0.1M HCl
NaCl + CO2 + H2O
[Na2CO3] = 0.0448 M
[NaHCO3] = 0.0704 M
total no. of mol of NaHCO3
= 2.8810-3 mol
 Original no. of mol of NaHCO3
= 2.8810-3 – 1.1210-3
= 1.76 10-3 mol
p. 171 Check Point 18-4
p.03
18.5 cm3 0.05M HCl: Phenolphthalein changes colour.
10.0 cm3 0.05M HCl: Methyl Orange changes colour.
25cm3 mixture containing
NaOH & Na2CO3 to be neutralized first
18.5 cm3 0.05M HCl
NaCl + NaHCO3
no. of mol of NaOH
= 9.2510-4 – 510-4
= 4.2510-4 mol
10.0 cm3 0.05M HCl
NaCl + CO2 + H2O
[NaOH] = 0.017 M
[Na2CO3] = 0.020 M
total no. of mol of NaHCO3
= 510-4 mol
 Original no. of mol of Na2CO3
= 510-4 mol
p.04
Equilibrium between (s) and (aq)
e.g.
sparingly soluble salt
(very low solubility in water!)
PbCl2(s)
“K” is very small!
Pb2+(aq) + 2Cl-(aq)
Ksp = [Pb2+(aq)] [Cl-(aq)]2
Key points :
1.
When PbCl2 is added to water, a very small amount of
ions will be formed. (as Ksp is very small)
2.
If extra ions (e.g. Cl-) are added to the solution, eqm will
shift BW, and more PbCl2(s) will be formed.
3.
If PbNO3(aq) and NaCl(aq) are mixed together, some of
Pb2+ and Cl- ions would form PbCl2(s). Conc. of ions
would decrease until Ksp is reached.
p.05
Solubility Product (Ksp)
and Solubility
Solubility of salt = conc. of salt dissolved (mol dm-3), or
= mass of salt dissolved per volume (g dm-3)
e.g. Given that Ksp of AgBr = 7.710-13 mol2 dm-6, calculate
the solubility of AgBr in g dm-3 in water.
7.710-13 = [Ag+][Br-]
 [Ag+] = [Br-] = 8.77 10-7 mol dm-3
 [AgBr] dissolved = 8.77 10-7 mol dm-3
 Solubility of AgBr = (8.77 10-7)(107.9+80.0) g dm-3
= 1.65 10-4 g dm-3
p.06
What is the solubility of AgBr in 0.001M NaBr(aq)?
AgBr(s)
at start
Ag+(aq) + Br-(aq)
0.001
X
at eqm X – a
a
0.001 + a
7.710-13 = a (0.001 + a)
a = 7.710-10
 [AgBr] dissolved = 7.7 10-10 mol dm-3
 Solubility of AgBr = (7.7 10-10)(107.9+80.0) g dm-3
= 1.45 10-7 g dm-3
Solubility of AgBr is reduced by “Common Ion Effect”.
p.07
Calculate Ksp from Solubility?
e.g. Given that solubilty of copper (I) bromide (CuBr)
= 2.010-4 mol dm-3, calculate the Ksp of CuBr.
CuBr(s)
at eqm X - 2.010-4
Cu+(aq) + Br-(aq)
2.010-4
2.010-4
Ksp = (2.010-4)2 = 4.010-8 mol2 dm-6
p.08
Predict “precipitation” with Ksp?
e.g. Given: Ksp of Ca(OH)2 = 8.010-6 mol3 dm-9.
If 2 dm3 0.20 M NaOH(aq) + 1 dm3 0.1M CaCl2(aq),
Will precipitation occur?
Ca(OH)2(s)
at start
0
Ca2+(aq) + 2OH-(aq)
0.033
0.133
ionic product = [Ca2+][OH-]2 = (0.033)(0.133)2
= 5.84 10-4 >> Ksp of Ca(OH)2
 Eqm will shift BW and precipitation will occur.
p.09
Assignment
Study all examples in p.173 - 176
p.179 Q.10, p.228 Q.18-20 [due date: 30/4(Thur)]
Pre-Lab: Expt. 14 Determination of Ksp of Ca(OH)2
Next ….
Redox Eqm [p.186 – 192]