Precipitates
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Precipitation Equilibria
• Involve sparingly soluble salts
•Consider a precipitate, AgCl
•AgCl ↔ (AgCl) ↔ Ag+ + ClKeq = Ksp = [Ag+][Cl-] = solubility product
Precipitation will not occur unless the
product of [Ag+] and [Cl-] exceeds Ksp
Ag2CrO4 ↔ 2 Ag+ + CrO42Ksp = [Ag+]2[CrO42-]
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Molar solubility
Equilibrium between solid compound and ions in a
saturated solution.
Molar solubility, S
No. of moles of a compound that is soluble in 1 liter
of saturated solution.
BaSO4(p)
Ba2+(aq) + SO42-(aq)
At ≡m,
SM
SM
Ksp = [Ba2+] [SO42-]= S2
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Solubility …
PbI2(p) ↔
Pb2+(aq) + 2I-(aq)
At ≡,
SM
2S M
[Pb2+] = S M
[I-] = 2S M
Ksp = [Pb2+] [I-] 2
= S (2S)2 = 4S3
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• AgCl(s)
Ag+ + Cl• Ksp= [Ag+][Cl-]
Spesies
Ag+
Cl-
initial
0M
0 M
∆
+x M
+x M
≡m
+x
+x
Ksp = x2 = S2 = 1.8 X 10-10
S =√ (1.8 X 10-10)
= 1.3 X 10-5 M
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• CaF2(s)
Ca2+ + 2F• Ksp =[Ca 2+ ][F-]2
species
Ca2+
F-
initial
0M
0M
∆
+x
+2x
≡m
+x
+2x
Ksp = (x)(2x)2 = 4 S3 = 1.5 X 10-10
S = 3.3 X 10-4 M
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The Common Ion Effect
common ion effect
• a salt will be less soluble if one of its
constituent ions is already present in the
solution
Calculate the molar solubility of lead iodide (a) in water
and (b) in 0.200 M sodium iodide solution. Ksp for PbI2
is 7.9 x 10-9
• (a) PbI2(s) ↔ Pb2+ + 2I•
•
•
•
•
•
Ksp = s(2s)2 = 4s3
= 7.9 x 10-9
s = 1.3 x 10-3 M
(b) Two sources of iodide: NaI and PbI2
Amount of iodide coming from PbI2 is small compared from NaI.
[I-] = CNaI + 2[Pb2+] ~ CNaI = 0.200 M
• S = Ksp = 7.9 x 10-9 = 2.0 x 10-7 M
•
[I-]2 (0.200)2
• Solubility has decreased (4 orders of magnitude) upon addition of excess
iodide.
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Effect of pH
• Solubility of a ppt whose anion is derived from
a weak acid will increase in the presence of
the added acid.
• MA ↔ M+ + A• A- + H+ ↔ HA
The anion can combine with the protons to
increase the solubility of the ppt.
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Effect of pH …
• Consider the solubility of CaC2O4 in the presence of
strong acid.
• CaC2O4 ↔ Ca2+ + C2O42- Ksp = [Ca2+] [C2O42-]
•
= 2.6 x 10-9
• C2O42- + H+ ↔ HC2O4- Ka2 = 6.1 x 10-5
• HC2O4- + H+ ↔ H2C2O4 Ka1 = 6.5 x 10-2
• Solubility s of CaC2O4 is equal to [Ca2+] = CH2C2O4
• Ksp = [Ca2+]CH2C2O4α2
• Ksp/ α2 = K’sp = [Ca2+]CH2C2O4 = s2
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• K’sp – conditional solubility constant
• α2 = _____Ka1Ka2________
•
[H+]2 + Ka1 [H+] + Ka1Ka2
• Calculate the solubility of CaC2O4 in a solution
containing 0.0010 M hydrochloric acid.
s2 = Ksp/ α2
• α2 = 5.7 x 10-2
• s = 2.1 x 10-4 M
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Effect of Complexation
• Ligands can compete for metal ion in a ppt.
(acids compete for anion)
• MA ↔ M+ + A• M+ + L ↔ ML+
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Consider the solubility of AgBr in the
presence of NH3
•
•
•
•
•
•
•
•
AgBr ↔ Ag+ + BrAg+ + NH3 ↔ Ag(NH3)+
Ag(NH3)+ + NH3 ↔ Ag(NH3)2+
Solubility s of AgBr is equal to [Br-] = CAg+
Ksp = [Ag+][Br-]
β0 = [Ag+]/CAg
Ksp = CAgβ0[Br-] = 4.0 x 10-13
Ksp / β0 = K’sp = CAg[Br-] = s2
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Calculate the solubility of silver thiocyanate in
(a) water, and (b) solution whose free ammonia
concentration is 0.0150 M.
• (a) In water
• AgSCN ↔ Ag+ + SCN•
Ksp = [Ag+][SCN-] = s2
•
s = √(1.1 x 10-12)
•
= 1.0 x 10-6 M
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(b) In the presence of ammonia
•
•
•
•
•
•
•
•
•
•
AgSCN ↔ Ag+ + SCNAg+ + NH3 ↔ Ag(NH3)+
Ag(NH3)+ + NH3 ↔ Ag(NH3)2+
s = [SCN-] = CAg
Ksp = CAgβ0 [SCN-] = β0s2
β0 = ___________1___________
1 + Kf1[NH3] + Kf1Kf2[NH3]2
= 2.66 x 10-4
1.1 x 10-12 = (2.66 x 10-4)s2
s = 6.4 x 10-5 N.Mohamed
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Separation by Precipitation
"A useful application of the solubility product is
the separation of one substance from another
by precipitating one from solution." This is the
basis of Qualitative Analysis.
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Precipitation Titrations
• Basis for ppt titration: ppt rxn must
– be rapid
– complete
– exist an indicator for detection of end-point
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• Frequently used reagent:
– AgNO3
– For the determination of halides, SCN- ,CN-, a few
divalent anions and mercaptans
•
M+
+
(titrant)
X- → MX(s)
(analyte)
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Titration Plot
• 50.00 mL of 0.100 M Cl- is titrated with
0.100 M AgNO3
• Plot pCl or pAg versus volume of AgNO3
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Titration Plot …
Addition of 0 mL titrant
pCl = -log Cl= -log (0.100)
= 1.000
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Before equivalence point
• Addition of 20 mL titrant
Cl+ Ag+
AgCl(p)
Initial 5.0
2.0
≡m 3.0 + x
x
2.0 - x
AgCl(p)
Ag+ + ClKsp = 1.0 x 10-10
Ignore x, Ksp very small
[Cl-] =
mmol Cl- not ppt
= 3.0 /70
total volume
= 0.043 M
• pCl = 1.37,
[Ag+] = 2.33 X 10-9 M
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• x boleh diabaikan kecuali terlalu hampir
kepada takat kesetaraan dimana [Cl-]
adalah kecil terutama sekali jika Ksp
adalah besar (10-4)
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Addition of 50 mL titrant: equivalence point
Initial
∆
≡m
Cl+
5.0 mmol
+x
x
Ag+
5.0 mmol
+x
x
AgCl(p)
-x
5.0 - x
Ksp = [Ag+][Cl-] = (x/100)2 = 1.0 x 10-10
(x/100) = [Cl-] = 1.0 x 10-5 M
pCl = 5.00
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Addition of 60 mL titrant: after equivalence
point
• mmol Ag+ in excess = 1.0 mmol
• Ksp = 1.0 [Cl-] = 1.0 x 10-10
110
[Cl-] = 1.1 x 10-8 M
pCl = 7.96
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• FAKTOR YANG MEMPENGARUHI
KETAJAMAN TAKAT AKHIR
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Locating End Points
•
•
•
•
•
Measure Cl- / Ag+ using suitable electrodes, or
Normally uses one of three different indicators
- potassium chromate
- dichlorofluorescein
- ferric nitrate
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Mohr’s Method
Developed for the determination of Cl- and BrUses AgNO3 as titrant and Na2CrO4 as indicator
Titration reaction:
Ag+ + ClAgCl(s)
Ksp = 1 x 10-10
•
white
Indicator reaction:
2Ag+ + CrO42- --> Ag2CrO4(s)
excess
yellow
red
• Ag2CrO4 is precipitated at the equivalence point
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Mohr’s Method ….
At the equivalence point: [Ag+] = [Cl-] = (Ksp)1/2 = 10-5
M
• Before the equivalence point: [Ag+] < 10-5 M.
• For formation of Ag2CrO4 precipitate,
[Ag+]2[CrO42-] = 1.1 x 10-12
(10-5)2[CrO42-] = 1.1 x 10-12
[CrO42-] = 1.1 x 10-2 M
Indicator reacts with added titrant
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Mohr’s Method ….
• Avoid masking of yellow from CrO42- ions to
the red of precipitated Ag2CrO4
• [CrO42-] is used from 0.002 - 0.005 M.
• Formation of Ag2CrO4 after equivalence point.
• A blank indicator is required.
• A solution containing indicator but no Cl- is
titrated with AgNO3 solution.
• → amount of Ag+ required to form Ag2CrO4
precipitate.
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Volhard’s Method
• For the determination of Ag+, Cl-, Br-, I-, SCNin acid medium.
• Add excess Ag+ to halide solution, followed by
back titration of the excess with thiocyanate.
• Fe3+ used as indicator – end-point marked by
appearance of red FeSCN2-(water soluble)
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Volhard’s Method …
• Analyte reaction:
X- + Ag+ (excess) --> AgX (s)
Titration reaction:
Ag+ + SCN- --> AgSCN (s)
white
Indicator reaction:
Fe3+ + SCN- --> Fe(SCN)2+
red
• For titrations of I-, Br-, SCN- - not necessary to separate AgX
precipitate before titrating with standard KSCN (AgX is less
soluble than AgSCN)
• However, for Cl- separation is required (react with titrant)
AgCl + SCN- --> AgSCN + ClN.Mohamed
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Fajan’s Method
• Use indicator that is adsorbed by ppt.
immediately after the equivalence point.
• Indicators have different colour in the free and
adsorbed form.
• E.g., dichlorofluorescein (dissociate to form H+
dan fluoresceinate anion).
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Fajan’s Method …..
• Titrate Cl- against AgNO3
• Before equivalence point, excess Cl•
AgCl: Cl- : Na+ (atau H+)
primary ion
counter ion
layer
layer
Indicator gives yellow-green colour to solution.
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Fajan’s Method …..
• After the equivalence point; Ag+ in excess.
•
AgCl: Ag+ : In• indicator is adsorbed on surface of ppt.
Fluoresceinate turns red on top of AgCl
precipitate.
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Fajan’s Method …..
• Use dextrin to avoid coagulation.
• Wish to obtain colloidal particles – high
surface area, so that quantity of indicator
adsorbed will be maximum.
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END
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KIDNEY
STONES
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What is a kidney
stone?
A kidney stone is a hard
mineral and crystalline
material formed within
the kidney or urinary
tract.
Kidney stones are a
common cause of pain in
the abdomen, flank, or
groin. Kidney stones
occur in 1 in 20 people at
some time in their life.
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The urinary tract
Consists of the kidneys, ureters, bladder, and
urethra. The kidneys remove extra water and
wastes from the blood, converting it to urine.
They also keep a stable balance of salts and
other substances in the blood.
Narrow tubes called ureters carry urine from
the kidneys to the bladder, an oval-shaped
chamber in the lower abdomen. Like a
balloon, the bladder's elastic walls stretch
and expand to store urine. They flatten
together when urine is emptied through the
urethra to outside the body.
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Kidney Stone ….
Kidney stone originates in the urinecollecting system of the kidney, and most
commonly causes pain when it travels down
the ureter to the bladder.
After traversing the bladder, it may enter the
urethra and continue to wreak havoc.
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Common Chemical Make Up:
calcium phosphate (8%)
calcium oxalate (most common: 73%, most opaque)
magnesium ammonium phosphate (also called "struvite" often caused by an infection)
Uncommon Chemical Make Up:
diammonium calcium phosphate
magnesium phosphate
Rare Chemical Make Up:
cystine (faintly opaque; 1%)
urate (lucent - meaning translucent to x-rays; 7%)
xanthine
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Who gets kidney stones?
Stones occur more frequently in men.
The prevalence of kidney stones rises
dramatically as men enter their 40s and
continues to rise into their 70s.
For women, the prevalence of kidney stones
peaks in their 50s.
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ROOT CAUSE
Metabolic disorders may be the root cause of
excessive calcium and oxalate forming stones in the
kidney.
The potential root causes are many and often complex,
include the following:
A re-absorption of the calcium from the bones back
into the blood system which the kidneys then filter out
(resorptive hypercalciuria or hyperparathyroidism).
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Root Cause ….
The intestines absorb too much calcium from
the diet (absorptive hypercalciuria).
The kidneys filter out calcium from the blood
but do not allow the reabsorption of the
calcium back into the blood as it should while
it is still in the tubule of the kidney (renal
hypercalciuria).
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ROOT CAUSE ….
Several forms of bowel disease
(ulcurative colitis, regional enteritis, etc.)
which can contribute to high levels of
urinary oxalate excretion.
Excess dietary intake of oxalate from
foods such as green leafy vegatables.
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Root Cause …..
High levels of uric acid in the urine can
promote calcium oxalate stones formation.
The lack of certain stone formation inhibitors
(eg citrate, magnesium) normally found in the
urine may not be present in sufficient
quantities and thereby allow the formation of
stones.
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END
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