CH. 17 ACID -- BASE EQUILIBRA & BUFFERS 17.1 Common ion effect 17.1 Calculation Henderson-...
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Transcript of CH. 17 ACID -- BASE EQUILIBRA & BUFFERS 17.1 Common ion effect 17.1 Calculation Henderson-...
CH. 17ACID -- BASE EQUILIBRA
& BUFFERS 17.1Common ion effect
17.1 CalculationHenderson- Hasselbalch eqnBuffers how works ion effect pH range
17.3Titration Curves indicators SA/SB - SA/WB WA/SB - Poly
Calculation pH
17.4SolubilityCalculation Ksp (Qsp) precipation
17.5Solubility; pHComplex Ions
17.6 & .7Ion GroupsSeparation
COMMON ION EFFECT
The shift in the position of an equilibrium on addition of a subst that provides an ion in common w/ one of the ions already involved in the equilibrium process.Also, decr solubility of soluble aqueous portion of ionic cmpd
Dissolve HNO2 in NaNO2 soln (H2O)
LeChatilier
add NO2-; from NaNO2
)aq(-12(aq)3 )l(2)aq(2 NO OH OH HNO
H3O+ + NO2-
H3O+ lowers pH; shifts away NO2- already formed
BUFFERSresist es in pHcomposed of: WA & CB or WB & CA
Add sm amt base to buffer soln Acidic component neutralize base added
BIO SYS: blood pH 7.4 control buffer of H2O/HCO3
-
Add sm amt acid to buffer soln Basic component neutralize acid added
HENDERSON -- HASSELBALCH
To prep buffer soln select WA w/ pKa close to 1 pH unit of buffer soln
How pH affects % dissoc WApH of buffer soln not depend on soln vol but on pKa & molar amt WA-CB
EX. Titrant 0.100 M NaOH and neutralize 40.0 ml of 0.100 M butanoic acid (HC4H7O2) Ka = 1.54*10-4
Find pH add 20.00 ml of titrant Find pH at equivalence pt & determine best indicator to use
Acid
BaseLog pK pH a
Base
Acid * K ]OH[ a3
Base
AcidLog- LogK- ]OH[Log- a3
1st determine # mmol HC4H7O2 present
mol HC4H7O2 = 40.0 mL * (0.100 mmol/1 mL) = 4.00 mmol need 4.00 mmol NaOH to reach equiv pt means need 40.0 ml of 0.100 M NaOH
b) Add 20.00 ml NaOH, also midpoint so pH = pKa moles WA: 4.00 mmol mol NaOH added: 20.00 ml*0.100 M= 2.00 mmol
2.0 mmol/60 ml = 0.0033 Mratio of [HC4H7O2]/[C4H7O2
-] = 0.033/0.033 = 1
a) No base, NaOH, added; butanoic WA: Ka = x2/[HA] x2 = (1.54*10-4)*(0.100) = .00392 pH = -Log(0.00392) = 2.41
3.81 0.00 3.81
033.0
0.033Log pK pH a
c) Add 40.0 ml NaOH @ equiv pt
All HC4H7O2 neutralized, find [C4H7O2-]
[C4H7O2-] = mmol/tot mL
pH = -Log(5.56*10-9) = 8.25
indicator: PHENOLPHTHALEIN
mmol C4H7O2- = 40.00 ml * (0.100 mmol/ml) = 4.0 mmol
[C4H7O2-] = 4.0 mmol/80.0 ml = 0.05 M
9-11
14-
-274b
w
11-4
14-
-274
Wb
10*5.56 )05.0)(10*49.6(
10*1
]OHC[*K
K ]H[
10*6.49 10*54.1
10*1
]OH[C
K K
soln basic anion WA,
Problem25.0-mL of 0.145 M HCl soln is titrated by 0.200 M KOH. How many mL of base must be added to reach the end point.
Want: mL KOH Data: 25.0-mL HCl @ 0.145 mol/L 0.200 mol/L KOH HCl + KOH --- H2O + KCl 1 mol HCl = 1 mol KOH
mol 0.200
L *
HCl mol 1
KOH mol 1 *
L
mol 0.145*HCl L 0.025 KOH mL-X
KOH mL-18.1 L
mL 1000 *L 018125.0
ProblemWhat is the pH at each point in the titration of 25.00-mL of 0.100 M HAcwith 0.100 M NaOH? a) before NaOH added b) after 10.00-mL c) after 12.50-mL d) after 25.00-mL (these are volumes added to original 25 ml)
a. Find initial [H3O+]; use Ka (1.8*10-5) HC2H3O2 + H2O - H3O+ + C2H3O2
-
Eq: 0.100-x x x
)10*8(0.100)(1. ][H x x-0.100
x K 5-
2
a = 1.3*10-3 pH = -log(1.3*10-3) = 2.89
b. 35-mL tot vol; amt neutralize= 0.025 L*(0.100 mol/L) = 0.0025 mol base added = 0.01 L*(0.100 M) = 0.001 mol
HC2H3O2 + OH - H2O + C2H3O2-
I 0.0025 0
C -0.001 +0.001
E 0.0015 0.001Convert to M using vol: .0015/.035 = .0429 M .001/.035 = .0286 M
4.56 0.18 - 4.74 0429.0
0286.0log 4.74
[HA]
][Alog pK pH
-
a
c. 37.5-mL tot vol; NaOH added=0.0125 L*(0.100 mol/L) = 0.00125 mol
HC2H3O2 + OH - H2O + C2H3O2-
I 0.0025 0
C -0.00125 +0.00125
E 0.00125 0.00125Convert to M using vol: .00125/.0375 = .0333 M .00125/.0375 = .0333 M
4.74 0.00 - 4.74 0333.0
0333.0log 4.74
[HA]
][Alog pK pH
-
a
d. Add 25-mL at eq.pt, neutralization complete; 50.00-mL tot vol amt C2H3O2
- present =0.0025 mol/0.050 L = 0.05 M
C2H3O2- + H2O - HC2H3O2 + OH-
Eq: 0.05-x x x Kb = Kw/Ka = 5.6*10-10
Kb = x2/0.05 x = 2.53*10-6 pOH = 5.28 then pH = 8.72
e. Follow up. What happens beyond the equivalence pt? pH is determined from excess base present. - add 26.0 mL, what is the pH? tot vol 51.00 mL
(0.026 L)*(0.100 M) = 0.0026 mol 0.0026 – 0.0025 = 0.0001 mol OH- excess
M = 0.0001/.051 = 1.96*10-3
11.3 2.7 - 14.0 pH 2.7 )10*log(1.96- pOH -3
Problem25.0-mL of 0.145 M HCl soln is titrated by 0.200 M KOH. How many mL of base must be added to reach the end point.
Want: mL KOH Data: 25.0-mL HCl @ 0.145 mol/L 0.200 mol/L KOH HCl + KOH --- H2O + KCl 1 mol HCl = 1 mol KOH
mol 0.200
L *
HCl mol 1
KOH mol 1 *
L
mol 0.145*HCl L 0.025 KOH mL-X
KOH mL-18.1 L
mL 1000 *L 018125.0
ProblemWhat is the pH at each point in the titration of 25.00-mL of 0.100 M HAcwith 0.100 M NaOH? a) before NaOH added b) after 10.00-mL c) after 12.50-mL d) after 25.00-mL (these are volumes added to original 25 ml)
a. Find initial [H3O+]; use Ka (1.8*10-5) HC2H3O2 + H2O - H3O+ + C2H3O2
-
Eq: 0.100-x x x
)10*8(0.100)(1. ][H x x-0.100
x K 5-
2
a = 1.3*10-3 pH = -log(1.3*10-3) = 2.89
b. 35-mL tot vol; amt neutralize=0.025L*(0.100 mol/L) = 0.0025 mol base added = 0.01 L*(0.100 M) = 0.001 mol
HC2H3O2 + OH - H2O + C2H3O2-
I 0.0025 0
C -0.001 +0.001
E 0.0015 0.001Convert to M using vol: .0015/.035 = .0429 M .001/.035 = .0286 M
4.56 0.18 - 4.74 0429.0
0286.0log 4.74
[HA]
][Alog pK pH
-
a
c. 37.5-mL tot vol; NaOH added=0.0125 L*(0.100 mol/L) = 0.00125 mol
HC2H3O2 + OH - H2O + C2H3O2-
I 0.0025 0
C -0.00125 +0.00125
E 0.00125 0.00125Convert to M using vol: .00125/.0375 = .0333 M .00125/.0375 = .0333 M
4.74 0.00 - 4.74 0333.0
0333.0log 4.74
[HA]
][Alog pK pH
-
a
d. Add 25-mL at eq.pt, neutralization complete; 50.00-mL tot vol amt C2H3O2
- present =0.0025 mol/0.050 L = 0.05 M
C2H3O2- + H2O - HC2H3O2 + OH-
Eq: 0.05-x x x Kb = Kw/Ka = 5.6*10-10
Kb = x2/0.05 x = 2.53*10-6 pOH = 5.28 then pH = 8.72
SOLUBILITY EQUILIBRA
Principals: examines quantitative aspects of solubility & ppt process
Ksp = solubility constant Cr2(SO4)3(s) --------> 2 Cr+3(aq) + 3 SO4-2(aq)
solid342
324
23
c)SO(Cr
SOCr Q
Qc + solid = Ksp
324
23sp SOCr K
Ksp
1. temp dependant 2. used to calc solubility of cmpd
Comparing Ksp
cmpds w/ same total # ions formed, then, higher Ksp is greater the solubility
Write dissoc eqn, # mols each ionWrite Ksp expressionFind Molarity - convert solubility to molar
CALCULATIONS Find Ksp
Solubility of silver dichromate @ 15oC is 8.3*10-3 g/100 mL
272
21sp OCrAg K Ag2Cr2O7(s) --------> 2 Ag+1(aq) + Cr2O7
-2(aq)
[Cr207-2] = 2 [Ag+]
0.0192 = 2*0.0192 = 0.0384
Find Solubility
Ksp = [0.0384]2 [0.0192] = 2.83*10-5
What is the molar solubility of iron III hydroxide in water? Ksp = 6.3*10-10
Write dissoc eqn, ion expression, ICE table
M 0.0192 g 431.8
mol 1*
L 1
mL 1000*
mL100
g 0.0083 M
Ksp = [Fe+3] [OH-]3
Ksp = [x] [3x]3 = 27x4
Ksp = 27x4
Fe(OH)3(s) --------> Fe+3(aq) + 3 OH-1(aq)
I ------ 0 0C -x +x +3xE -x x 3x
M10*2.20 27
10*6.3
27
K x 3-4
-104
sp
Ignore: solid
PPT FORMATION -- will or not form ???
Qsp = Ksp soln satur, no Qsp< Ksp soln unsat, no pptQsp > Ksp ppt form till soln satur
Will ppt form, if so, what is it?0.20 L of 0.050 M Na3PO4 & 0.10 L of 0.20 M Ca(NO3)2
Ions present: Na+1 PO4-3 Ca+2 NO3
-1
solubility rules: Ca3(PO4)2 (s) the ppt, insoluble
Ca3(PO4)2 (s) <----> 3 Ca+2 (aq) +2 PO4-3 (aq)
Ksp = [Ca+2]3[PO4-3]2
Ksp = 1.2*10-29
Qsp = [Ca+2]3 [PO4-3]2
= (0.067)3(0.033)2 = 3.28*10-7
mols Ca+2 = 0.10 L * 0.20 M = 0.020 mols[Ca+2] = 0.020 mols/0.30 L = 0.067 M
Qsp > Ksp, so ppt will form Ca3(PO4)2 ppt till Qsp = 1.2*10-29
mols PO4-3 = 0.20 L * 0.05 M = 0.10 mols
[PO4-3] = 0.10 mols/0.30 L = 0.033 M
If Qsp = 6.7*10-35 What then??? Qsp < Ksp, no ppt
COMPLEX IONS
Central METAL ion covalent
bonded to 2+ LIGANDS
Metal ion acts as Lewis Acid,Ligand acts as Base
Can separate metal from its ore, isolate & remove toxic metal, convert metal ion to diff form
ION GROUPS
Separation of 2 competing ppting metal ions thru diff properties of solubility of cmpds
Add ppting ion till Qsp of more soluble ion close to its Ksp
This allows max amt of less soluble cmpd ppt out, while none of the more soluble
Mixture of many diff metal ions, use various techniques to sep ions into characteristic groups pg 737
Group 1 Insol Chloridesppt w/ HCl Ag+ Hg2
+2 Pb+2
Group 2 Acid-insol Sulfidesppt w/ H2SCu+2 Cd+2 As+3 Sb+3 Bi+3
Sn+2 Sn+4 Hg+2 Pb+2
Group 3 Base-insol S-2 & OH-1
ppt w/ NH4+-NH3 buffer
Zn+2 Mn+2 Ni+2 Fe+2 Co+2 as S-2
Al+3 Cr+3 as OH-
Group 4 Insol PO4-3
ppt w/ (NH4)2HPO4, NaCO3
Mg+2 Ca+2 Ba+2
Group 5 Alkali & NH4+ Ions
generally speaking, what’s left Use various techniques to ppt out specific ions from each group
Calculate the pH of a soln that is 0.350 M pyridinium chloride (C5H5NHCl) and 0.210 M pyridine (C5H5N).
WB C5H5N & C5H5NHCl contain common ion effect: C5H5NH+
I 0.210 0.350 0.0 C -x +x +x E 0.210-x 0.350+x x
C5H5N(aq) ) + H2O(l) C5H5NH+(aq) + OH-(aq)
assume 5% rule for WA
0.210
0.350x 10*1.7
210.0
x)(x)(0.350
N]H[C
]NHH][C[OH 9-
55
55
x = [OH-] = 1.02*10-9 pOH = -Log(1.02*10-9) = 8.99 pH = 14 – 8.99 = 5.01
Kb = 1.7*10-9
The addition of bromide ion will decrease the water solubility of which of the following salts?
a.BaSO4 b. Li2CO3 c. PbS d. AgBrAgBr
Which pair of compounds will form a buffer solution when dissolved in water in equimolar amounts?
a. HCl and KCl b. HNO3 and NaNO3
c. HCl and NH4Cl d. NH3 and NH4Cl
NH3 and NH4Cl
The Ka of HCN is 4.9 x 10-10. What is the pH of a buffer solution that is 0.100 M in both HCN and KCN?
a. 4.7 b. 7.0 c. 9.3 d. 14.09.3 same concentrations,-log(4.9*10-10)
The Ka of HCN is 4.9 x 10-10. What is the pH of a buffer solution that is 0.100 M in HCN and 0.200 M in KCN?
a. 7.0 b. 9.0 c. 9.3 d. 9.69.6
pH = pKa + log[KCN]/[HCN]
The Ka of HCN is 4.9 x 10-10. What is the pH of a buffer solution that is 1.00 M in HCN and 0.100 M in KCN?
a. 7.0 b. 8.3 c. 9.0 d. 9.38.3 pH = pKa + log[KCN]/[HCN]
In titrating a weak base with a strong acid, the best indicator to use would be:
a. methyl red (changes color at pH = 5).
b. bromothymol blue (changes at pH = 7).
c. phenolphthalein (changes at pH = 9).
d. none of the above. methyl red (changes color at pH = 5)
Titrating a weak acid with a strong base, best indicator to use would be:a. methyl red (changes color at pH = 5).
b. bromothymol blue (changes at pH = 7).
c. phenolphthalein (changes at pH = 9).
d. none of the above.
phenolphthalein (changes at pH = 9)
The Ksp of BaCO3 is 5.0 x 10-9. What is the concentration of barium ion in a saturated aqueous solution of BaCO3?
a. 7.1 x 10-5 M b. 2.5 x 10-9 M c. 5.0 x 10-9 M d. 1.0 x 10-8 M
7.1 x 10-5 M 5.0*10-9 = x*x5.0*10-9 = x2
The Ksp of BaF2 is 1.7 x 10-6. What is the concentration of barium ion in a saturated aqueous solution of BaF2?
a. 1.7 x 10-6 M b. 3.4 x 10-6 M c. 7.5 x 10-3 M d. 1.5 x 10-2 M
7.5 x 10-3 M
Ksp = [Ba+2][F-]2
= (x)*(2x)2
1.7*10-6 = 4x3
The Ksp of BaF2 is 1.7 x 10-6. What is the concentration of fluoride ion in a saturated aqueous solution of BaF2?
a. 1.7 x 10-6 M b. 5.7 x 10-5 M c. 7.6 x 10-3 M d. 1.5 x 10-2 M
1.5 x 10-2 M
BaF2 -- Ba+2 + 2 F-
[F-] =2x = 2(7.5*10-3)
Which of the following substances will be more soluble in acidic solution than in basic solution:(a)Ni(OH)2(s), (b) CaCO3(s), (c) BaF2(s), (d) AgCl(s)? So, which is more soluble at low pH than high pH
Ni(OH)2 more soluble in acidic, basicity of OH-
CaCO3 dissolves in acidic, as CO3- basic anion
BaF2 dissolves, as F- basic anion
(d) The solubility of AgCl is unaffected by changes in pH because Cl– is the anion of a strong acid and therefore has negligible basicity.
Addition of _____ will increase the solubility of MgCO3.
MgCO3(s) Mg2+(aq) + CO32–(aq)
• MgCl2
• Na2CO3
• NaOH
• HCl
• KHCO3
HCl
Predict the order of precipitation of Ba2+, Pb2+, Ca2+ with the addition of NaF.
• Ca2+ then Ba2+ then Pb2+
• Pb2+ then Ba2+ then Ca2+
• Pb2+ then Ca2+ then Ba2+
• Ca2+ then Pb2+ then Ba2+
• Ba2+ then Pb2+ then Ca2+
Ksp
BaF2 1.7 x 10–6
PbF2 3.6 x 10–8
CaF2 3.9 x 10–11
Ca2+ then Pb2+ then Ba2+
Predict the order of precipitation of Cl–, CO32–, Br– upon the
addition of AgNO3.
• Br– then CO32– then Cl–
• Br– then Cl– then CO32–
• Cl– then CO32– then Br–
• Cl– then Br– then CO32–
• CO32– then Cl– then Br–
Ksp
AgCl 1.8 x 10–10
Ag2CO3 8.1 x 10–12
AgBr 5.0 x 10–13
Br– then Cl– then CO32–
A solution contains 1.0 × 10-2 M Ag+ and 2.0 ×10-2 M Pb2+. When Cl– is added to the solution, both AgCl (Ksp = 1.8× 10-10) and PbCl2 (Ksp = 1.7×10-5) precipitate from the solution. What concentration of Cl– is necessary to begin the precipitation of each salt? Which salt precipitates first?
We are given Ksp values for the two possible precipitates. Using these and the metal ion concentrations, we can calculate what concentration of Cl– ion would be necessary to begin precipitation of each. The salt requiring the lower Cl– ion concentration will precipitate first.