C. Y. Yeung (CHW, 2009)

p.01

Acid-Base Eqm (7): DoublDouble Indicator Titration e Indicator Titration & Solu& Solubility Product (Kbility Product (Kspsp))

For mixtures containing For mixtures containing TWO BASESTWO BASES. (e.. (e.g. NaOH / Nag. NaOH / Na22COCO33 / NaHCO / NaHCO33))

Double Indicator TitrationDouble Indicator Titration[Phenolphthalein & Methyl Orange][Phenolphthalein & Methyl Orange]

decreasing Kdecreasing Kbb

When titrate against standard HCl(aq):When titrate against standard HCl(aq):

NaOH + HCl NaOH + HCl NaCl + H NaCl + H22OO

NaNa22COCO33 + HCl + HCl NaHCO NaHCO33 + NaCl + NaCl

NaHCONaHCO33 + HCl + HCl NaCl + CO NaCl + CO22 + H + H22OO

p.02

E.g.E.g.

25cm25cm33 mixture containing mixture containing NaHCONaHCO33 & Na & Na22COCO33

11.2 cm11.2 cm33 0.1M HCl: 0.1M HCl: PhenolphthaleinPhenolphthalein changes colour. 2 changes colour. 28.8 cm8.8 cm33 0.1M HCl: 0.1M HCl: Methyl OrangeMethyl Orange changes colour. changes colour.

to be neutralized firstto be neutralized first

11.2 cm11.2 cm33 0.1M HCl 0.1M HCl

NaHCONaHCO33

28.8 cm28.8 cm33 0.1M HCl 0.1M HCl

NaCl + CONaCl + CO22 + H + H22OO

no. of mol of Nano. of mol of Na22COCO33

= 1.12= 1.121010-3-3 mol mol

total no. of mol of NaHCOtotal no. of mol of NaHCO33

= 2.88= 2.881010-3-3 mol mol

Original no. of mol of NaHCOOriginal no. of mol of NaHCO33

= 2.88= 2.881010-3-3 – 1.12 – 1.121010-3-3

= 1.76 = 1.76 1010-3-3 mol mol[Na[Na22COCO33] = 0.0448 M] = 0.0448 M

[NaHCO[NaHCO33] = 0.0704 M] = 0.0704 M

All COAll CO332-2- converted to HCO converted to HCO33

--..

All HCOAll HCO33-- converted to CO converted to CO22 and H and H22O.O.

p. 171 Check Point 18-4p. 171 Check Point 18-4p.03

25cm25cm33 mixture containing mixture containing NaOH & NaNaOH & Na22COCO33

18.5 cm18.5 cm33 0.05M HCl: 0.05M HCl: PhenolphthaleinPhenolphthalein changes colour. changes colour. 10.0 cm10.0 cm33 0.05M HCl: 0.05M HCl: Methyl OrangeMethyl Orange changes colour. changes colour.

to be neutralized firstto be neutralized first

10.0 cm10.0 cm33 0.05M HCl 0.05M HCl

NaCl + CONaCl + CO22 + H + H22OO total no. of mol of NaHCOtotal no. of mol of NaHCO33

= 5= 51010-4-4 mol mol

no. of mol of NaOH no. of mol of NaOH = 9.25= 9.251010-4-4 – 5 – 51010-4-4 = 4.25= 4.251010-4-4

molmol

Original no. of mol of NaOriginal no. of mol of Na22COCO33

= 5= 51010-4-4 mol mol[NaOH] = 0.017 M[NaOH] = 0.017 M

[Na[Na22COCO33] = 0.020 M] = 0.020 M

18.5 cm18.5 cm3 3 0.05M HCl 0.05M HCl

NaCl + NaHCONaCl + NaHCO33

p.04

Equilibrium between Equilibrium between (s)(s) and and (aq)(aq)

e.g.e.g.

PbClPbCl22(s)(s) PbPb2+2+(aq) (aq) + + 22ClCl--(aq)(aq)

sparingly soluble salt sparingly soluble salt (very low solubility in water!)(very low solubility in water!) ““K” is very small!K” is very small!

KKspsp = [Pb = [Pb2+2+(aq)] [Cl(aq)] [Cl--(aq)](aq)]22

Key points :Key points :1.1. When PbClWhen PbCl22 is added to water, a very small amount of ion is added to water, a very small amount of ion

s will be formed. (as Ks will be formed. (as Kspsp is very small) is very small)

2.2. If extra ions (e.g. ClIf extra ions (e.g. Cl--) are added to the solution, eqm will s) are added to the solution, eqm will shift BW, and more PbClhift BW, and more PbCl22(s) will be formed.(s) will be formed.

3.3. If PbNOIf PbNO33(aq) and NaCl(aq) are mixed together, some of Pb(aq) and NaCl(aq) are mixed together, some of Pb2+2+ and Cl and Cl-- ions would form PbCl ions would form PbCl22(s). Conc. of ions would d(s). Conc. of ions would d

ecrease until Kecrease until Kspsp is reached. is reached.

p.05

Solubility Product (KSolubility Product (Kspsp))

and and SolubilitySolubilitySolubility of salt Solubility of salt = conc. of salt dissolved (mol dm= conc. of salt dissolved (mol dm-3-3), or), or

= mass of salt dissolved per volume (g dm= mass of salt dissolved per volume (g dm-3-3))

e.g.e.g. Given that KGiven that Kspsp of AgBr = 7.7 of AgBr = 7.71010-13-13 mol mol2 2 dmdm-6-6, calculate th, calculate th

e solubility of AgBr in g dme solubility of AgBr in g dm-3-3 in water. in water.

7.77.71010-13-13 = [Ag = [Ag++][Br][Br--]]

[Ag[Ag++] = [Br] = [Br--] = 8.77 ] = 8.77 1010-7-7 mol dm mol dm-3-3

[AgBr] dissolved = 8.77 [AgBr] dissolved = 8.77 1010-7-7 mol dm mol dm-3-3

Solubility of AgBr = (8.77 Solubility of AgBr = (8.77 1010-7-7)(107.9+80.0) g dm)(107.9+80.0) g dm-3-3

= 1.65 = 1.65 1010-4-4 g dm g dm-3-3

p.06

What is the solubility of AgBr in 0.001M NaBr(aq)?What is the solubility of AgBr in 0.001M NaBr(aq)?

AgBr(s)AgBr(s) AgAg++(aq) (aq) + + BrBr--(aq)(aq)XXat startat start 0.0010.001

at eqmat eqm X – aX – a aa 0.001 + a0.001 + a

7.77.71010-13-13 = a (0.001 + a) = a (0.001 + a)

a = 7.7a = 7.71010-10-10

[AgBr] dissolved = 7.7 [AgBr] dissolved = 7.7 1010-10-10 mol dm mol dm-3-3

Solubility of AgBr = (7.7 Solubility of AgBr = (7.7 1010-10-10)(107.9+80.0) g dm)(107.9+80.0) g dm-3-3

= 1.45 = 1.45 1010-7-7 g dm g dm-3-3

Solubility of AgBr is reduced by “Common Ion Effect”.Solubility of AgBr is reduced by “Common Ion Effect”.

p.07

Calculate KCalculate Kspsp from Solubility? from Solubility?

e.g.e.g. Given that solubilty of copper (I) bromide (CuBr) Given that solubilty of copper (I) bromide (CuBr) = 2.0= 2.01010-4-4 mol mol dmdm-3-3, calculate the K, calculate the Kspsp of CuBr. of CuBr.

KKspsp = (2.0 = (2.01010-4-4))22 = 4.0 = 4.01010-8-8 mol mol22 dm dm-6-6

CuBr(s)CuBr(s) CuCu++(aq) (aq) + + BrBr--(aq)(aq)

at eqmat eqm X - 2.0X - 2.01010-4-4 2.02.01010-4-4 2.02.01010-4-4

p.08

Predict “precipitation” with KPredict “precipitation” with Kspsp??

e.g.e.g. Given: KGiven: Kspsp of Ca(OH) of Ca(OH)22 = 8.0 = 8.01010-6-6 mol mol3 3 dmdm-9-9. .

If 2 dm If 2 dm33 0.20 M NaOH(aq) + 1 dm 0.20 M NaOH(aq) + 1 dm33 0.1M CaCl 0.1M CaCl22(aq), (aq),

Will precipitation occur?Will precipitation occur?

Ca(OH)Ca(OH)22(s)(s) CaCa2+2+(aq) (aq) + + 2OH2OH--(aq)(aq)

at startat start 0.1330.13300 0.0330.033

ionic product = [Caionic product = [Ca2+2+][OH][OH--]]22 = (0.033)(0.133) = (0.033)(0.133)22

= 5.84 = 5.84 1010-4 -4 >> K>> Kspsp of Ca(OH) of Ca(OH)22

Eqm will shift BW and precipitation will occur.Eqm will shift BW and precipitation will occur.

AssignmentAssignment

p.09

Next ….Next ….Redox Eqm [p.186 – 192]Redox Eqm [p.186 – 192]

Study all examples in p.173 - 176Study all examples in p.173 - 176

p.179 Q.10, p.228 Q.18-20 p.179 Q.10, p.228 Q.18-20 [due date: 30/4(Thur)] [due date: 30/4(Thur)]

Pre-Lab: Expt. 14 Determination of KPre-Lab: Expt. 14 Determination of Kspsp of Ca(OH) of Ca(OH)22