Chapter 9
Aqueous Solutions and Chemical Equilibria
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CaCO322 (s ) CO (gl ) + H O ( ) (limestone) (water falls) 2- Ca (aq ) 2HCO3 ( aq )
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Classifying Solutions of Electrolytes
Electrolytes—solutes form ions when dissolved in water (or certain other solvents, e.g. acetonitrile)
Strong (weak, non-) electrolytes
-+ HNO32 + H O NO 3 + H 3 O (completely ionized) -+ CH3233 COOH+ H O CH COO + H O (partially ionized) sugar (s) + water = sugar aquesous soln (completely non-ionized)
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1 HF
2+ - - Ca(OH)2 Ca +2OH , no Ca(OH) in solution HF is a weak electrolyte due to hydrogen bond.
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Acids and Bases
An acid donates protons, a base accepts protons An acid donates protons only in the presence of a proton acceptor (a base). A base accepts protons only in the presence of a proton donor (an acid)
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Bronsted-Lowry Acids and Bases
These two chemists pointed out that acids and bases can be seen as proton transfer reactions.
According to the Bronsted-Lowry concept: • An acid donates a proton and the base accepts it. • Newly formed species are called conjugate base and conjugate acid, respectively.
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2 acid + base conjugate base + conjugate acid
- HCl + NH34 Cl + NH
Acid12 + base base 1 + acid 2
A conjugate base is formed when an acid loses a proton (HClCl-); A conjugate acid is formed + when a base accepts a proton (NH3NH4 )
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Consider the reaction of NH3 with H2O:
+ - NH3 (aq) + H2O (aq) NH4 (aq) + OH (aq) (base1) (acid2) (acid1) (base2)
Note:
+ NH3 & NH4 are a conjugate acid-base pair. - H2O & OH : also a conjugate acid-base pair
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Example
Identify the acid and base species in the following equations:
2- - - CO3 (aq) + H2O(l) HCO3 (aq) + OH
-+ CH3233 COOH + H O CH COO + H O (Acetic acid) (base) (hydronium)
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3 Amphiprotic species (“amphi-”: both):
a species that can act as either an acid or a base, depending on the other reactant.
Consider water:
- - H2O + CH3O OH + CH3OH acid base
+ - H2O + HBr H3O + Br
base acid
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-- HPO24 + HO 2 HPO 34 + OH base1 acid2 acid1 base2 ()amphiprotic -2-+ H24 PO + H 2 O HPO 4 + H 3 O acid1 base2 base1 acid2
+- NH22 CH COOH NH 3 CH 2 COO glycine zwitterion “sweeterian” A zwitterion is an ion that bears both a positive and a negative charge
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AUTOPROTOLYSIS In autoprotolysis a proton is transferred between two identical molecules, one of which acts as a Brønsted acid, releasing a proton which is accepted by the other molecule acting as a Brønsted base.
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4 AUTOPROTOLYSIS
Water undergoes self-ionization autoprotolysis,
since H2O acts as an acid and a base.
+ - H2O + H2O H3O + OH The extent of autoprotolysis is very small.
The equilibrium constant expression for this reaction is:
+ - [H3O ] [OH ] Kc = 2 [H2O] The concentration of water is essentially constant.
2 + - Therefore: [H2O] Kc = [H3O ] . [OH ]
constant = Kw
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Example: Calculate the hydronium and hydroxide ion concentration of pure water at 25 ◦C and 100 ◦C.
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We call the equilibrium value of the ion product + - [H3O ][OH ] the ion-product constant of water.
+ - -14 Kw = [H3O ] [OH ] = 1.0 x 10 at 25°C
+ Using Kw you can calculate concentrations of H3O and OH- in pure water. + - -14 [H3O ] [OH ] = 1.0 x 10 + - But [H3O ] = [OH ] in pure water + - -7 [H3O ] = [OH ] = 1.0 x 10 M
If you add an acid or a base to water the concentrations + - of H3O and OH will no longer be equal. But Kw will still hold.
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5 Acid / Base Strengths
The strength of an acid or base is determined by the extent it dissociates in water to form H+ or OH- respectively
Strong acids: HA H+- + A (100% dissociation) + [H ] = [HA]initial
Strong bases: MOH M+- + OH (100% dissociation) [OH ] = [MOH]initial
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Weak Acid
Weak acids only partially dissociate to give H+ in H2O Equilibrium Constant + - HA + H2O H3O + A
+ - Ka = [H ][A ] / [HA]
Ka – acid dissociation constant
The smaller the Ka value, the weaker the acid
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Weak Base
Weak bases react with water by abstracting a proton Equilibrium Constant + - B + H2O BH + OH
+ - Kb = [BH ][OH ] / [B]
Kb – base hydrolysis constant Hydrolysis – reaction with water
The smaller the Kb value, the weaker the base
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6 • The strongest acids have the weakest conjugate bases and the strongest bases have the weakest conjugate acids. • A acid (base) can be cationic, anionic, neutral.
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Differentiating/Leveling Solvents
In water, perchloric and hydrochloric acids are strong acids and have the same strength.
In acetic acid, perchloric acid is considerably stronger than HCl; its dissociation being ~5000 times greater.
Acetic acid acts as differentiating solvent toward the two acids by revealing the inherent differences in their activities, and water is called leveling solvent for perchloric, hydrochloric (and nitric) acids because all three are 100% ionized in water and show no difference in strength.
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Equilibrium Constant
aA bB... cC dD ... [][]...CDcd K [AB ]ab [ ] ...
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7 Standard State
1. Solution concentration ([X]) unit: M (mol/L). 2. Gas concentrations (partial pressure): atm. 3. The concentrations of pure solids, pure liquids, and solvents are omitted because they are (constant) unity. 4. K = f (T). 5. Only thermodynamic (tendency) rather than kinetic (reaction rate) conditions are considered.
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Calculation of Ka or Kb from its Conjugate Kb or Ka
[H+- ][A ] Acid: HA H+- + A K a [HA] Conjugate base of HA acid, A [HA][OH- ] A + H O HA + OH- K 2 b [A- ] [H+- ][A ] [HA][OH - ] KK = [H+- ][OH ] K ab[HA] [A- ] w
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8 Manipulating Equilibrium Constants
(1) AB CD K1 (2) EF GH K2
(3) (1) (2) KKK312 (3)ABEF CDGH K3 ?
(4) (1) (2)KKK412 / (4)ABGH CDEF K4 ?
[Class practice] 25
Solubility Product Constant
(Ksp) In General: (For sparingly soluble compounds) +y -x AxBy (s) xA + yB
[A+y]x [B-x]y K = ------[AxBy]
+y x -x y [AxBy]K=Ksp =[A ] [B ]
{[AxBy](solid)isconstant}
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Examples of Ksp Expression
AgCl() s Ag Cl Ksp [ Ag ][ Cl ] 22 2 CaF2 () s Ca2 F Ksp [ Ca ][ F ] 22 22 BaSO44() s Ba SO Ksp [ Ba ][ SO 4 ] 22 22 CaCO33() s Ca CO Ksp [ Ca ][ CO 3 ] 2 PbI2 () s Pb2? I Ksp
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9 Calculating Ksp from the Solubility
If the solubility of AgCl is 1.9x10-3 g/L, what
would be the Ksp?
+- AgCl( s ) Ag Cl , Ksp = [Ag ][Cl ] (sss ) +- 2 52 10 Kssp = [Ag ][Cl ] (1.3 10 ) 1.7 10 [1.910/sgLM3 ? 1.910/33gL 1.910/ gL = 1.3 105 M ] FW. . AgCl 143.4 g / mol
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The solubility of lead(II) arsenate, Pb3(AsO4)2, is -5 3.0x10 g/L, calculate its Ksp.
23 2332 Pb342( AsO ) ( s ) 3 Pb 2 AsO 4sp K = [ Pb ] [ AsO 4 ] (sss ) 3 2 23 32 3 2 5 36 KPbAsOsssp = [ ] [ 4 ] (3 ) (2 ) 108 s 4.2 10 [3.010/sgLM5 ? 3.0 1055gL / 3.0 10 gL / = 3.3 108 M ] F.. W Pb342 ( AsO ) 899.44 gmol /
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Calculating Solubility from Ksp AgI() s Ag I (sss ) 2 Ksp [][]Ag I s s s
sK sp
2 PbI2 () s Pb 2 I (sss ) 2 22 23 Ksp [][]Pb I =(2) s s = 4 s KK s 3 sp() sp 1/3 44 30
10 A general expression for a precipitate dissociation
Axy B (s) xy A + B (sxsys ) xy x y xyxy() KABxsysxyssp [][] [ ][ ] KK1 ss()xy sp ; ( sp ) xy xyxy xy xy
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Solubility of a Precipitate in Pure Water
How many grams of AgCl (fw = 143.32) can be dissolved in 100.00 mL of water at o -10 25 C? (Ksp [AgCl] =1.82X10 ) AgCl() s Ag Cl (sss ) 2 KAgClssssp =[ ][ ]= 1/2 10 1/2 sK()sp (1.8210) 1.82 1055 1.35 10 M
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. How many grams of AgCl (fw = 143.32) can be dissolved in 100.00 mL of water at 25 oC? [from previous calculation: s =1.35X10-5 M]
weight(# of moles ) formula weight (CV ) formulaweight ( C s ) = [1.35 1053 (M ) 100.00 10 (Lgmol )] 143.32( / ) = 2.0 10-4 g = 0.20 mg
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11 The Common Ion Effect
In AgCl ( s ) Ag Cl system, add a strong electrolyte containing Ag+- or Cl ions (the same as the dissociated ions--common ions) eg.., NaCl Na Cl or AgNO33Ag NO
. In the presence of common ions, the solubility of the precipitate will dramatically decrease— Common Ion Effect. [Le Chatelier’s Law]
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Calculate the solubility of slightly soluble barium
fluoride, BaF2, in (a) pure water, and (b) 0.15 M -6 NaF. (Ksp = 1.0x10 ) (a) In pure water
2 BaF2 () s Ba 2 F (sss ) 2 22 23 Ksp [][](2)4Ba F s s s K 1.0 106 s ()sp 1/3 ( ) 1/3 44 6.3 103 M
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(b) In 0.15 M NaF
2 BaF2 ( s ) Ba 2 F ; NaF Na F (0.15 M ) (s ) ssF 2 [ ]remaining [ F ] remaining 0.15 M 22 2 KBaFssp [ ][ ] (0.15) 0.0225 s K 1.0 106 sMsp 4.4 105 0.0225 0.0225 Check the assumption: 2s + [F-] = 8.8x10-5 + 0.15 = 0.15 M Assumption was valid.
BaF2 solubility in 0.15 M NaF is ~1/140 times that in pure water.
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12 Class Practice: Calculate the molar
solubility of Ag2CO3 in a solution that is -12 0.0200 M in Na2CO3. (Ksp =8.1X10 )
(key: 1.0x10-5 M)
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. EXAMPLE: Calculate the molar solubility
of Ag2CO3 that results when 200. mL of 0.0100 M AgNO3 is mixed with 100. mL of -12 0.100 M Na2CO3. Ksp = 8.1 X 10
The chemical reaction :
2AgNO32323 Na CO Ag CO ( s ) 2 NaNO 3 moles 2 1 1 2 0.0100 200. 0.100 100. 0 0 initial moles 1000 1000
0.002 0.01 (2:10 molar ratio, Na CO in excess) 23 11 0 0.01- (0.002) ( 0.002 ) 0.002 At eq. moles 2 2
0.009 0.001 0.00 2
2- # of moles 0.009 [CO ] = = 0.030 M 3 solution volume in L 0.200 0.100
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222 Ag23 CO( s ) 2 Ag CO 3 Ksp [ Ag ] [ CO 3 ] 22 (sssCOCO ) 2 [33 ]remaining [ ] remaining 22 2 12 KAgCOssp [ ] [3 ] (2 ) (0.030) 8.1 10 8.1 1012 sM ( )1/2 8.2 10 6 0.030 4
Check the assumption : 22 6 [CO33 ] s [ CO ]remaining 8.2 10 0.030 0.030 M Assumption was valid !
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13 Precipitation Calculations
? xy A + B Axy B (s)
initial conc . [ A ]ii [ B ] 0
x y As Ax B y (s) xy A + B KAB sp [ ] eq [ ] eq
Solubility quotient (Qc ) (or ion product) x y QAci [ ] [B]i
If QKcsp , precipitation occures (saturated solution)
If QKcsp , NO precipitation (unsaturated solution)
If QKcsp , the reaction mixture is at equilibrium
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2+ 2- Suppose a solution has [Ca ] and [SO4 ] of 0.0052 M and 0.0041 M, respectively. If these concentrations were doubled by evaporating half the water in the solution,
would precipitation of CaSO4 occur given that the Ksp for calcium sulfate is 2.4 x 10-5? Solution:
22? Ca SO44 CaSO
22 QCaSOc [ ][4 ] (0.0052 2)(0.0041 2) -5 =8.83 10 Ksp You would expect precipitation to occur.
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A solution of 0.00016 M lead(II)nitrate, Pb(NO3)2, was poured into 456 mL of 0.00023 M sodium sulfate,
Na2SO4. Would a precipitate of lead(II)sulfate, PbSO4, be expected to form if 255 mL of the lead nitrate
solution were added? (The Ksp for lead(II)sulfate is 1.7 -8 x10 ) 22? Pb SO44 PbSO Solution:
()CV 2 ()CV 2 22 Pb SO4 QPbSOc [][]4 VVtotal total (0.00016MmL 255 ) (0.00023 465) = (456 255)mL (456 255) mL =5.74 10-5MM 1.48 10 -4 ) -9 =8.48 10 Ksp You would NOT expect precipitation to occur. 42
14 Completeness of Precipitation
Lead chromate, PbCrO4, is a yellow pigment used in paints. -5 Suppose 0.50 L of a 1.0x10 MPb(C2H3O2) and 0.50 L of a -3 1.0x10 MK2CrO4 solution are mixed. Calculate the equilibrium concentration of Pb2+ ion remaining in the
solution after PbCrO4 precipitates. What is the percentage of Pb2+ remaining in solution after the precipitation has occurred. -14 (The Ksp for PbCrO4 is 1.8x10 ).
2+ 2- Pb + CrO44 PbCrO (s) Ini. # moles 0.50 1.0 10-5 0.50 1.0 10 -3 0 = 5.0 10-6 = 5.0 10 -4 0 Final. # moles 0 5.0 10-4 - 5.0 10-6 5.0 10 -6 Final. Conc. (M) 0 4.95 10-4 5.0 10 -6
ie. ., solid PbCrO4 is in contact with a solution -4 2 containing 4.95 10 M CrO4 ions. 43
2+ 2- PbCrO44 (s) Pb + CrO [Initial] 0 4.95 10-4 [At e.q.] (s) s 4.95 10-4 + s 4.95 10 -4
2+ 2- -4 -14 Ksp [Pb ][CrO4 ] = s (4.95 10 ) = 1.8 10 s = [Pb2+ ] = 3.6 10 -11 M 3.6 10-11 % of Pb2+ in solution vs precipitated: 100% = 7.3 10-4 % 5.0 10-6
A complete precipitation!
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Fractional Precipitation
Fractional precipitation is the technique of separating two or more ions from a solution by adding a reactant that precipitates first one ion, then another, and so forth.
2+ 2+KCrO24 2+ (Ba , Sr ) BaCrO4 ( Sr )
KCrO24 2+ 2+ SrCrO4 (when [Ba ] [Sr ]) [KK (2.3 10-10 (4 10 -5 )] sp[BaCrO44 ] sp [SrCrO ]
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15 Separation by Precipitation
EXAMPLE: Can Fe3+ and Mg2+ be separated quantitatively as hydroxides from a solution that is 0.10 M in each cation? If the separation is possible, what range of OH- concentrations is permissible.
3+ - 3 -39 Ksp =[Fe ][OH ] =2X10
2+ - 2 -12 Ksp =[Mg ][OH ] =7.1X10
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Assume quantitative separation requires that the concentration of the less soluble material to have decreased to < 1x10-6 M before the more soluble material begins to precipitate. Assume [Fe3+] = 1.0x10-6 M What will be the [OH-] required to reduce the [Fe3+]to[Fe3+] = 1.0x10-6 M ? 3+ - 3 -39 Ksp =[Fe ][OH ] = 2x10 2x10-39 [OH-]3 = ------= 2x10-33 1.0x10-6 [OH-] = 1.3x10-11 M
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2+ - 2 -12 Ksp =[Mg ][OH ] = 7.1 X 10
What [OH-] is required to begin the precipitation
of Mg(OH)2? [Mg+2] = 0.10 M (0.10 M)[OH-]2 = 7.1 X 10-12
7.1x10-12 [OH-]2 = ------= 7.1x10-11 0.10 [OH-] = 8.4x10-6 M
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16 Since 1.3x10-11 Mislessthan8.4x10-6 M, the Fe3+ will be quantitatively removed
long before the Mg(OH)2 begins to form. Therefore, the separation is possible between [OH-]1.3x10-11 Mand[OH-]= 8.4x10-6 M. (pH range?)
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Effect of pH on Solubility
2 CaF2 () s Ca 2 F AgCl() s Ag Cl AgCN() s Ag CN
What changes in their solubilities if (1) + HCl (2) + HNO -+ 3 F + H HF (weak acid) (3) H SO 24 CN-+ + H HCN (weak acid) 2+ 2- Ca + SO44 CaSO (s)
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17 Which of the following does affect the solubility of a precipitate ?
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A saturated solution of Cu(II) hydroxide has a pH of 10. Calculate
the Ksp for the Cu(II) hydroxide.
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Acid-Base Dissociation Constants
Example
+ - Calculating concentrations of H3O and OH in solutions of a strong acid or base.
Calculate the concentrations of hydronium ion and hydroxide ion at 25°C in 0.10 M HCl.
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18 + [H3O ] Calculation of Weak Acids
[H O+- ][A ] HA + H O H O+- + A K = 3 23 a[HA]
cHA ~0 0 [][]xx cxHA x xK a = [ cxHA ] 22 KcaHA [ x ] = x or x + KxKc a aHA 0 KK2 4 Kc x [H O+ ] aa aHA 3 2
If KccxaHAHAHA very small and/or large : c + x [H3a O ] K cHA (if cKHAa / 1000, Err% 1.6% )
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[OH-] Calculation of Weak Bases
[BH+- ][OH ] B + H O BH+- + OH K = 2b[B]
cB 0 ~0 [][]xx cxBb x x K = [ cxB ] 22 KcxxbbbB [B ] = or xKxKc + 0 KK2 4 Kc x [OH- ] bb bB 2
If KccxcbBBB very small and/or large : -+- xKcK[OH]bB ; [HO]= 3 w /[OH]
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19 Buffer Solutions
A buffer is a solution characterized by the ability to resist changes in pH with dilution or with addition of limited amounts of acid or base.
Buffers contain either a weak acid and its conjugate base or a weak base and its conjugate acid. Thus, a buffer solution contains both an acid species and a base species in equilibrium.
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Strong base Strong acid neutralizes weak acid neutralizes weak base
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How do we calculate the pH of a buffer ?
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20 Consider mixture of salt NaA and weak acid HA.
+ - NaA (s) Na (aq) + A (aq) [H+][A-] Ka = HA (aq) H+ (aq) + A- (aq) [HA]
K [HA] Henderson-Hasselbalch [H+] = a [A-] equation Take log [HA] -log [H+] = -log K -log a - [conjugate base] [A ] pH = pKa + log [A-] [acid] -log [H+] = -log K + log a [HA] Assumption: Changes in [A-] & [HA] Remember pX = -logX will be negligible (within 5%) if so pK = -log K a a Ka < 0.01 & Ka < 0.01 [A-] [base] [acid] pH = pKa + log [HA] use initial conc. of acid and base to calculate pH 61
“Ladder Diagram”
Acid–base ladder diagram for (a) acetic acid showing the relative concentrations of CH3COOH and _, CH3COO and (b) Carbonic acid.
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What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?(Potassium formate)
Mixture of weak acid and conjugate base!
HCOOH (aq) H+ (aq) + HCOO- (aq) Initial (M) 0.30 0.00 0.52 Change (M) -x +x +x Equilibrium (M) 0.30 - x x 0.52 + x [HCOO-] Common ion effect pH = pK + log a [HCOOH] 0.30 – x 0.30 [0.52] 0.52 + x 0.52 pH = 3.77 + log = 4.01 [0.30]
HCOOH pKa = 3.77
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21 Which of the following are buffer systems? (a) KF/HF
(b) KBr/HBr, (c) Na2CO3/NaHCO3. (d) NaHCO3
(a) HF is a weak acid and F- is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution 2- - (c) CO3 is a weak base and HCO3 is its conjugate acid buffer solution
(d) ? Amphiprotic species
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Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?
+ + NH4 (aq) H (aq) + NH3 (aq)
[NH ] [0.30] pH = pK + log 3 pK = 9.25 pH = 9.25 + log a + a = 9.17 [NH4 ] [0.36]
-10 Ka = 5.6 x 10
pOH = 14 - pH = 4.73 [OH-] = 1.8 x 10-5 M
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Calculate moles of all components. Remember we have
80.0 mL of the buffer solution @ 0.30 M NH3/0.36 M NH4Cl and 20.0 mL of 0.050 M NaOH.
+ Moles NH4 = (0.080 L)(0.36 M) = 0.029 moles Moles NH3 = (0.080 L)(0.30 M) = 0.024 moles Moles NaOH = (0.020 L)(0.050 M) = 0.0010 moles final volume = 80.0 mL + 20.0 mL = 100 mL start (moles) 0.029 0.0010 0.024 + - NH4 (aq) + OH (aq) H2O (l) + NH3 (aq) end (moles) 0.028 0.0 0.025 Moles/Volume(L)
0.028 0.025 [0.25] [NH +] = [NH ] = pH = 9.25 + log = 9.20 4 0.10 3 0.10 [0.28] ΔpH = 0.03
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22 Property of Buffer Solutions
The pH remains essentially independent of dilution. Very small changes in pH when limited amounts of acid or base are added (pH <0.01)
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How do we build a better buffer? Add approximately equal quantities of acid and base Have relatively high concentrations of acid and base the larger the [acid] & [base] the greater the buffer capacity
How do we prepare a buffer at a given pH? Choose acid/base conjugate pair from table Check to be sure that they are unreactive in the system used
pKa pH typical rule of thumb
pH + 1 = pKa
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Create buffer with pH = 7.50.
pH + 1 = pKa
Look for pKa‘s 6.5 8.5 -7 -9 => Ka‘s 3 x 10 3 x 10
- -8 HOCl / OCl Ka = 3.5 x 10 - -2 -8 H2PO4 / HPO4 Ka2 = 6.2 x 10 OK? - -2 -8 H2AsO4 / HAsO4 Ka2 = 8 x 10 - -7 H2CO3 / HCO3 Ka1 = 4.3 x 10
Calculate quantities of acid and base. [conjugate base] pH = pK + log a [acid]
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23 - -7 H2CO3 / HCO3 Ka1 = 4.3 x 10
pKa1 = 6.37
- [HCO3 ] pH - pKa = log [H2CO3] 7.50 - 6.37 = 1.13
[HCO -] 10-1.13 = 13.6 = 3 [H2CO3]
Set [H2CO3] = 0.0100 M (NOTE: This is a judgment call.)
- then [HCO3 ] = 13.6 [H2CO3] = 0.136 M
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Chapter 9 Summary
• Classification of electrolytes • Common-ion effect • Equilibrium constants • Acids and bases • Ion-product of water • Acid-base dissociation • Conjugate acid-base pairs • Buffer capacity • Buffer solutions • Amphiprotic species • Autoprotolysis • Solubility product
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24 Important Equations [Y]yz [Z] For wx W + X yzK Y + Z = [W]wx [X] +- +- 2H23 O H O + OH K w3 = [H O ][OH ] +- +- or H2w O H + OH K = [H ][OH ] Weak acid HA [H+- ][A ] HA H+- + A KKc = [H + ] = aaHA[HA] Weak base NaA +- - - NaA Na + A A+ HO2 HA + OH - [HA][OH ] - Kw KKccbbNaANaA = - [OH ] = = [A ] Ka Buffer solution--Henderson-H asselbalch equa tion weak acid /congj. base or weak b ase/congj. acid
[]Base cNaA pH = plKa og = plogKa []Acid cHA 73
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