Biology/Chemistry 302 -- Biochemistry Aqueous equilibria and buffer construction
. (last revised 2013 Sep 12). Goals: Making solutions. 1) Given a periodic table, ...
[email protected]. Standard buffer solutions (20°C). We offer a broad range
of solutions from pH 1 to pH 13 (20°C). Product. Composition. Capacity. Cat. no.
ammonia R1. Dilute to 100.0 ml with water R. Phosphate buffer solution pH 3.5. 4000700. Dissolve 68.0 g of potassium dihydrogen phosphate R in water R and ...
Buffer solutions have the useful property of resisting changes in pH when hydrogen ions are added to or subtracted from the solution by chemical reactions or by ...
adjust the pH (2.2.3) with phosphoric acid R. Sulphate buffer solution pH 2.0. 4008900. Dissolve 132.1 g of ammonium sulphate R in water R and dilute to 500.0 ...
3.40 g of potassium dihydrogen phosphate R in water R ... Sulphate buffer solution pH 2.0. ... To 250.0 ml of 0.2 M potassium hydrogen phthalate R add.
Apr 6, 1977 - Department ofPhysics, Guy's Hospital Medical School, London SE1 9RT, U.K. ... pH changes during the freezing of buffer solutions containing ...
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Von: D:\REZ-EM\EM-TEM anal\PUFFER & rel\Preparation of pH buffer solutions(AnalChem resources,ex web10-04-16mht).mht. Preparation of pH buffer ...
80 ml. 4x Upper gel buffer. 0.5 M Tris-Cl, pH 6.8,. 0.4% SDS. 2. 121.1. 70-80. 80 ml. 4.1 10% SDS. 1L: 100g SDS into 1 L, heat to 68 o. C for solubility. pH ~6.6.
10 Jul 2011 ... Larutan penyangga adalah larutan yang bersifat mempertahankan pH-nya, jika
ditambahkan sedikit asam atau sedikit basa atau diencerkan.
2 L. 484 g Tris. 114.2 ml glacial acetic acid. 200 ml 0.5 M EDTA 8.0. To make 1x TAE 20 L, add 400 ml 50X buffer into 19.6 L ddH2O. 4. SDS-PAGE Gel Solutions.
Page: 1. pH and Buffers. pH. The negative logarithm of the hydrogen ion ...
commonly called the 'buffer equation', which relates the pH of solution to the.
Buffer Preparation (Gozani Lab). 1. 1 M Tris-HCl Buffers. pH. Volume (L). TrisBase (g). HCl (ml) ... Add 40 g sucrose to 50 ml 0.04% BPB solution, adjust final volume 100 ml. 7. SDS-PAGE ... 100 ml 10x TBS. 10 ml 10% Tween20 (final 0.1% v/v).
0.5L of 1M K2HPO4 at 174.18g mol-1 = 87.09g. 0.5L of 1M KH2PO4 at 136.09g mol-1 = 68.045g preparation of 0.1 M potassium phosphate buffer at 25°C.
Information from cshprotocols.org: Gomori buffers, the most commonly used phosphate buffers, consist of a mixture of monobasic dihydrogen phosphate and ...
August 28, 2009 [PROBLEM SET FROM R. CHANG TEST BANK] www.kau.edu.
sa/aasiri2 2. 5. Calculate the pH of a buffer solution that contains 0.25 M benzoic
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Konstantinos Yiannopoulos, Kyriakos G. Vlachos, Member, IEEE, and Emmanouel Varvarigos. AbstractâWe present an architecture for implementing optical.
28 Jun 2012 ... The write buffer in Spansion® MirrorBit® Flash memory devices is designed to ...
The page buffer read function accelerates read operations.
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In GIS, we can classify buffer primitives as point buffering operations, line
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Poshdar, M., González, V.A., and Kasiviswanathan, B. (2018). âBuffer Management in Construction - A. New Zealand Study.â In: Proc. 26th Annual Conference of ...
Information from cshprotocols.org: Gomori buffers, the most commonly used phosphate buffers, consist of a mixture of monobasic dihydrogen phosphate and ...
Phosphates inhibit many enzymatic reactions and procedures that are the foundation of molecular cloning ... 0.5L of 1M K2HPO4 at 174.18g mol-1 = 87.09g.Missing:
Buffer = chemical system that resists changes in pH. Example. Add 0.010 mol of
HCl or NaOH to: Buffer Solutions. pH. 1 L Solution. Initial after HCl after NaOH.
Buffer Solutions John W. Moore Conrad L. Stanitski Peter C. Jurs http://academic.cengage.com/chemistry/moore
Chapter 16 Additional Aqueous Equilibria Stephen C. Foster • Mississippi State University
Buffer Action
Buffers are made with ~equal quantities of a conjugate acid-base pair. (e.g. CH3COOH + CH3COONa)
Buffer Action
1 L Solution
Initial
pH after HCl
after NaOH
pure H2O
7.00
2.00
12.00
[CH3COOH] = 0.5 M + [CH3COONa] = 0.5 M
4.74
4.72
4.76
A buffered solution
Added OH- removed by the acid: CH3COOH + OHCH3COO- + H2O Added H3O+ removed by the conjugate base: CH3COO- + H3O+ CH3COOH + H2O The equilibrium maintains the acid/base ratio. CH3COOH + H2O CH3COO- + H3O+ pH remains stable.
The pH of Buffer Solutions
Blood pH • Blood is buffered at pH = 7.40 ± 0.05. pH too low: acidosis. pH too high: alkalosis. • CO2 generates the most important blood buffer.
H2CO3(aq) + H2O(ℓ)
Example Add 0.010 mol of HCl or NaOH to:
Buffer Action
Buffers must contain: • A weak acid to react with any added base. • A weak base to react with any added acid. • These components must not react with each other.
CO2(aq) + H2O(ℓ)
Buffer = chemical system that resists changes in pH.
H2CO3(aq) H3O+(aq) + HCO3-(aq)
Depends on the [acid]/[base] – not absolute amounts. HendersonHenderson-Hasselbalch equation [H3O+] = Ka
What is the pH of a 0.050 M (monoprotic) pyruvic acid + 0.060 M sodium pyruvate buffer? Ka = 3.2 x 10-3.
pH
weak acid
weak base
Ka(weak acid) pKa
4 lactic acid
lactate ion
1.4 x 10-4
3.85
5 acetic acid
acetate ion
1.8 x 10-5
4.74
6 carbonic acid
hydrogen carbonate ion
4.2 x 10-7
6.38
= 2.49 + 0.08
7 dihydrogen phosphate hydrogen phosphate ion
6.2 x 10-8
7.21
= 2.57
8 hypochlorous acid
hypochlorite ion
3.5 x 10-8
7.46
9 ammonium ion
ammonia
5.6 x 10-10
9.25
carbonate ion
4.8 x 10-11 10.32
pH = – log(3.2 x
10-3)
+ log(0.060/0.050)
What is the HPO42-/H2PO4- ratio in blood at pH =7.40. Ka(H2PO4- ) = Ka,2(H3PO4) = 6.2 x 10-8. 7.40 = −log(6.2 x 10-8) + log([HPO42-]/[H2PO4-]) log([HPO42-]/[H2PO4-]) = 0.192 [HPO42-]/[H2PO4-] = 1.5
Addition of Acid or Base to a Buffer 1.0 L of buffer is prepared with [NaH2PO4] = 0.40 M and [Na2HPO4] = 0.25 M. Calculate the pH of: (a) the buffer (b) after 0.10 mol of NaOH is added. Ka (H2PO4-) = 6.2 x 10-8
pH = pKa + log
[A-] [HA]
pH = 7.21 + log 0.25 = 7.01 0.40
Buffer Capacity The amount of acid (or base) that can be added without large pH changes. [A-]/[HA] determines the buffer pH. Magnitude of [A-] and [HA] determine buffer capacity The buffer capacity for: • Base addition = nconjugate acid • Acid addition = nconjugate base
Useful buffer range: pH = pKa ±1 (10:1 or 1:10 ratio of [A-]/[HA]).
Addition of Acid or Base to a Buffer 1.00 L Buffer: [NaH2PO4]= 0.40 M ; [Na2HPO4] = 0.25 M. (b) calculate pH after 0.10 mol of NaOH is added. Ka (H2PO4-) = 6.2 x 10-8
(b) 0.10 mol of NaOH, converts conj. acid to base: H2PO4- + OH0.40 0.10 (0.40 – 0.10) 0
ninitial nadded nafter
pKa = -log(6.2 x 10-8) = 7.21 (a) No base added:
10 hydrogen carbonate
pH = pKa + log
[A-] [HA]
→ HPO42- + H2O 0.25 (0.25 + 0.10)
Use n directly. [ ] = n/ V and V is the same for both (cancels)
pH = 7.21 + log 0.35 = 7.28 0.30
Acid-Base Titrations • Standard solution (titrant titrant) is added from a buret. • The equivalence point occurs when a stoichiometric amount of titrant has been added. • Use a pH meter. • An indicator is used to find an end point. point • Color change observed. • End point ≠ equivalence point (should be close...). ntitrant = nanalyte
Detection of the Equivalence Point Acid-Base Indicator = weak acid that changes color with changes in pH. HIn(aq) + H2O(ℓ) color 1
H3O+(aq) + In-(aq) color 2
Observed color will vary (depends on [HIn] and [In-] in solution). Ka =
Detection of the Equivalence Point The acidic form dominates when the [HIn] >> [In-] If
pH = pKa -1
Basic color shows when [In-] >> [HIn] [In-] = 10 [HIn]
Ka =
- pKa = 1 - pH
Detection of the Equivalence Point
[H3O+][In-] [H3O+] = [HIn] 10
Ka =
- pKa = - pH - 1
If
[H3O+][In-] [HIn]
[HIn] = 10 [In-]
[H3O+][In-] = 10 [H3O+] [HIn]
pH = pKa +1
Titration of Strong Acid with Strong Base AcidAcid-base titration curve = plot of pH vs Vtitrant added. Titrate 50.0 mL of 0.100 M HCl with 0.100 M NaOH
Titration of Strong Acid with Strong Base Vtitrant/mL Vexcess/mL Vtotal/mL
[OH-]/mol L-1
Titration of Strong Acid with Strong Base
pH
50.0
0
100.0
0
7.00
50.1
0.1
100.1
0.0001
10.00
50.2
0.2
100.2
0.0002
10.30
51
1
101.0
0.0010
11.00
55
5
105.0
0.0048
11.67
60
10
110.0
0.0091
11.96
70
20
120.0
0.0167
12.22
80
30
130.0
0.0200
12.30
0.1 mL of base increased the pH by 3 units!
Titration of Weak Acid with Strong Base
Titration of Weak Acid with Strong Base
More complicated. • Weak acid is in equilibrium with its conjugate base. Example Titrate 50.0 mL of 0.100 M acetic acid with 0.100 M NaOH. What is the pH after the following titrant additions: 0 mL, 40.0 mL, 50.0 mL, and 50.2 mL?
40.0 mL added
0 mL added
HA + H2O
Ka = 1.8 x 10-5 =
2 [H3O+][A-] ≈ x [HA] (0.100)
50 mL of 0.100 M
ninitial nadded nleft
HA(aq) + OH-(aq)
+
H2O(ℓ)
0.00500 0.00400 0.00100
0.00400
Each OH- removes 1 HA… nleft = 0.00500 - 0.00400
H3O+ + A-
A-(aq)
…and makes 1 A40 mL of 0.100 M
x= 0.0013 Vtotal = (50.0 + 40.0) mL = 90.0 mL = 0.0900 L
pH = -log(0.0013) = 2.88
Titration of Weak Acid with Strong Base
Titration of Weak Acid with Strong Base
40.0 mL added
(c) 50.0 mL added Equivalence. All HA is converted to A-.
Henderson-Hasselbalch: pH = -log(1.8 x 10-5) + log
[A-] = nA-/ V
(0.00400/0.0900) (0.00100/0.0900) [HA] = nHA/ V
HA(aq) + OH-(aq) ninitial nadded nleft
A-(aq)
+
H2O(ℓ)
0.00500 0.00500 0
0.00500
Note: V cancels (could be omitted)
pH = 4.74 + log
0.0400 = 5.34 0.0100
Vtotal= 100.0 mL, so [A-] = 0.00500 mol/0.100 L = 0.0500 M A- is basic!
4
Titration of Weak Acid with Strong Base
Titration of Weak Acid with Strong Base
Use Kb to solve for the [OH-] generated by [A-]:
(c) 50.2 mL added 0.2 mL of “extra” OH- dominates pH. Ignore any contribution from A-.
Kb =
[OH-][HA] [A-]
A+ (0.0500 - x)
Kb = Kw / Ka = 5.6 x 10-10 H2 O
Kb = 5.6 x 10-10 ≈ pOH = 5.28
HA x
x2 0.0500
+
OHx
x = 5.3 x 10-6 M
[OH-] = (0.0002 L x 0.100 mol/L) / 0.1002 L = 2.0 x 10-4 M pOH = 3.7 pH = 14.0 – 3.7
= 10.3
pH = 14 - pOH = 8.72
Titration of Weak Acid with Strong Base
Titration of Weak Base with Strong Acid
Titrate of 50.0 mL of 0.100 M acetic acid with 0.100 M NaOH.
Volume of 0.100 M HCl added (mL)
pH = pKa at 50% titration (25 mL) pKa (acetic acid) = 4.74. Short vertical section compared to strong acid/strong base.
Solubility Equilibria and KSP Some ionic compounds are slightly water soluble. Saturation occurs at low concentration. AgCl(s)
50.0 mL of 0.100 M ammonia titrated with 0.100M HCl. • pH = pKa at 50% to equivalence (pKa = 9.25).
Solubility Equilibria and KSP Example Ksp(AgCl)=1.8 x 10-10. Calculate the solubility (mol/L).
Ag+(aq) + Cl-(aq) Ksp = [Ag+][Cl-] = 1.8 x 10-10
The solubility product constant is: Ksp = [ Ag+] [ Cl-]
At equilibrium,
[Ag+] = [Cl-] = S
Then:
1.8 x 10-10 = (S)(S) = S2 S = 1.3 x 10-5 M
As always, [ ] of solid is omitted
Solubility = 1.3 x 10-5 mol/L.
5
Factors Affecting Solubility
Common Ion Effect
pH and Dissolving Slightly Soluble Salts Using Acids Insoluble salts containing anions of Bronsted-Lowry bases dissolve in acidic solutions.
AgCl is slightly soluble in water:
Ca2+(aq) + CO32- (aq) Ksp = 8.7 x 10-9
Pure water: CaCO3(s) Acid:
Add more Ag+ or Cl-. Equilibrium will move left.
Called the common ion effect. CO3
HCO3
Ag+(aq) + Cl-(aq)
AgCl(s)
-(aq)
2-(aq)
+ H3
+ H3
O+(aq)
O+(aq)
H2CO3(aq)
HCO3
-(aq)
+ H2O(ℓ)
H2CO3(aq) + H2O(ℓ) CO2(g) + H2O(ℓ)
K ≈ 105
e.g. AgCl is less soluble in a NaCl solution than in water. The “common ion” is Cl-.
Solubility and the Common Ion Effect
Solubility and the Common Ion Effect
Example Calculate the solubility of PbI2 in (a) water (b) 0.010 M NaI. Ksp (PbI2) = 8.7 x 10-9.
(b) NaI supplies I - lowering the PbI2 solubility.
Ksp = 8.7 x 10-9 = [Pb2+][I -]2 = S(2S + 0.010)2 Ksp = [Pb2+][I -]2 = (S)(2S)2 = 4S3 = 8.7 x 10-9 S = 1.3 x
10-3
M
Assume S Ksp • Q must decrease. • remove ions, precipitate solid.
If Q = Ksp • at equilibrium (saturated solution).
If Q < Ksp • Q must increase. • dissolve more solid (if present). Form more ions.
Precipitation: Will It Occur?
Precipitation: Will It Occur?
Mix 25.0 mL of 0.0025 M HCl and 10.0 mL of 0.010 M AgNO3. Will AgCl precipitate? Ksp (AgCl) = 1.8 x 10-10.
Slowly add HCl(aq) to a solution that is 0.010 M in Cu+ and 0.500 M in Pb2+. Which salt will precipitate first? Ksp for PbCl2 and CuCl are 1.7 x 10-5 and 1.9 x 10-7.
nCl-
= 0.0250 L (0.0025 mol/L)
nAg+ = 0.0100 L (0.010 mol/L)
= 6.25 x 10-5 mol = 1.0 x 10-4 mol
CuCl: KSP= 1.9 x 10-7 = [Cu+][Cl-] = 0.010[Cl-]
Vtotal = (0.0250 + 0.0100) = 0.0350 L. [Cl-] = 6.25 x 10-5 mol/ 0.0350 L = 1.79 x 10-3 mol/L [Ag+]
= 1.0 x
10-4
mol / 0.0350 L = 2.86 x
10-3
mol/L
Q = [Ag+][Cl-] = (1.79 x 10-3)(2.86 x 10-3) = 5.1 x 10-6 Q > KSP
Find the minimum [Cl-] that will cause precipitation:
[Cl-] = 1.9 x 10-5 M
Least soluble, precipitates 1st
PbCl2: KSP = 1.7 x 10-5 = [Pb2+][Cl-]2 = 0.500[Cl-]2 [Cl-] = 5.8 x 10-3 M
