Solubility – the extent at which a solid will dissolve into a particular
solvent
Our focus today: The solubility of salts in aqueous solutions
Solubility Rules
1. All common compounds of Group I and
ammonium ions are soluble.
2. All nitrates, acetates, and chlorates are soluble.
3. All binary compounds of the halogens (other than
F) with metals are soluble, except those of Ag,
Hg(I), and Pb. Pb halides are soluble in hot water.)
4. All sulfates are soluble, except those of barium,
strontium, calcium, lead, silver, and mercury (I).
The latter three are slightly soluble.
5. Except for rule 1, carbonates, hydroxides, oxides,
silicates, and phosphates are insoluble.
6. Sulfides are insoluble except for calcium, barium,
strontium, magnesium, sodium, potassium, and
ammonium.
Applying the rules
Ca(OH)2
Pb(NO3)2
BaSO4
KCl
Solubility is relative. There is really a continuum
Soluble, very slightly soluble, slightly soluble, insoluble, very insoluble
Can we be more quantitative about solubility?
Solubility Product Constant
Solids are not included in the equilibrium expression. This is
because the concentration (mass/volume) of a solid is its density, which
is a constant. Since it remains constant throughout the course of a
reaction it is incorporated into the equilibrium constant.
Example:
PbCO3(s)  Pb2+(aq) + CO32-(aq)
Ksp = [Pb2+][CO32-] = 7.4×10-14
Experiment 1: Add some solid lead carbonate into a beaker of water.
What will be the equilibrium concentrations of Pb2+ and CO32-?
[Pb2+] must equal [CO32-], because their only source is from PbCO3(s)
7.4*10-14 = [Pb2+][CO32-] = [Pb2+]2 = x2
x = [Pb2+] = [CO32-] = 2.7*10-7 M
What is it solubility?
Determine the solubility of magnesium hydroxide (Ksp = 1.5×10-11)
Mg(OH)2 ↔ Mg2+ + 2OHKsp = [Mg2+][OH-]2 = (x)(2x)2 = 4x3 = 1.5×10-11
x = 1.6×10-4 M
Determine the solubility of Zinc phosphate (Ksp = 9.1×10-33)
Sr3(PO4)2 ↔ 3Sr2+ + 2PO43Ksp = [Sr2+]3[PO43-]2 = (3x)3(2x)2 = 36x5 = 9.1×10-33
X = 1.9×10-7 M
Experiment 2: Add some lead carbonate and 0.1 mol of sodium carbonate into a
beaker and add water until volume is 1.00 L.
Ksp = [Pb2+][CO32-] = [Pb2+](0.1 + [CO32-]lead carb)
Let x = [Pb2+], [Pb2+] = [CO32-]lead carb, so
Ksp = x(0.1+x) = 7.4*10-14
It is best to make the assumption that 0.1 >>> x, and then check to make
sure that it is according to your answer.
Ksp = x(0.1) = 7.4*10-14
x = [Pb2+] = 7.4*10-13 M, assumption was valid
This illustrates the common ion effect!!!!
Addition of sodium carbonate at the treatment plant to the municipal
water supply dramatically reduces the level of soluble lead that can
come from lead plumbing.
Forming a precipitate from combining two soluble salts
0.05 M
2KI(aq)
+
0.1 M
Pb(NO3)2(aq) ↔ 2K+ + 2NO3- + PbI2(s)
Do we form a precipitate?
Ksp(PbI2)
= 7.1×10-9
Determine the final [I-] and [Pb2+]
2I-(aq) +
0.05 M
0.05 – 2x
Ksp
Pb2+(aq)
0.1 M
0.1-x
↔
PbI2(s)
x
= 7.1×10-9 = [Pb2+][ I-]2 = (0.1-x)(0.05-2x)2
x = 0.02484 mol precipitate/L solution
[Pb2+] = 0.0752 M
[I-] = 3.232×10-4 M
How about if we add 1 mM of each? Will a precipitate form?
1 mM
2KI(aq)
+
1 mM
Pb(NO3)2(aq) ↔ 2K+ + 2NO3- + PbI2(s)
solubility (expressed as mol/L, g/L, ppm (mg/L), and of salt dissolved
Determine the solubility of CaCO3 in 0.2 M Na2CO3 in ppm
Ksp(CaCO3) = 3.8×10-4
The effect of Complex Ion Formation
For any real system there are many reactions that are at equilibrium. All
equilibrium conditions must be satisfied simultaneously (very important
concept!!!!!)
Example1:
Place lead iodide in water
*Without considering other soluble lead iodide species,
we can say 2[Pb2+] = [I-], and let x =[Pb2+]
Ksp = x(2x)2 = 4x3 , x = [Pb2+] = 0.001255 M
However, Life is not always this simple!!
Write the pertinent reactions.
PbI2(s) ↔ Pb2+ +2I-
Ksp = 7.9*10-9 = [Pb2+][I-]2
Pb2+ + I- ↔ PbI+(aq)
K1 = 100 =[PbI+(aq)]/[Pb2+][I-]
Pb2+ + 2I- ↔ PbI2(aq)
2 = 1400 = [PbI2(aq)]/[Pb2+][I-]2
Pb2+ + 3I- ↔ PbI3-(aq)
3 = 8300 = [PbI3-(aq)]/[Pb2+][I-]3
Pb2+ + 4I- ↔ PbI42-(aq)
4 = 30000 = [PbI42-(aq)]/[Pb2+][I-]4
[Pb2+] is the same in all six of these equilibrium expressions. There
can only
be one [Pb2+] and one [I-] concentration.
So, how does one tackle this problem? There are five equations and six
unknowns. We need more independent equations.
__________________________________________________________
______
Define soluble lead – any soluble species containing lead contributes to
the total
concentration of soluble lead.
Soluble lead = [Pb2+] + [PbI+(aq)] + [PbI2(aq)] + [PbI3-(aq)] + [PbI42-(aq)]
And then there is mass balance,
If we let x = mol Pb atoms, 2x = mol I atoms
2(mol Pb atoms) = mol I atoms
(mass balance equation)
Summary
Ksp = 7.9*10-9 = [Pb2+][I-]2
K1 = 100 =[PbI+(aq)]/[Pb2+][I-]
2 = 1400 = [PbI2(aq)]/[Pb2+][I-]2
3 = 8300 = [PbI3-(aq)]/[Pb2+][I-]3
4 = 30000 = [PbI42-(aq)]/[Pb2+][I-]4
2[soluble lead] = [I-] + [PbI+(aq)] + 2[PbI2(aq)] + 3[PbI3-(aq)] +
24[PbI4 (aq)]
2(mol Pb atoms) = mol I atoms
So,
Using Excel and successive approximations (more on this in a little bit)
[I-] =
0.0023971 M
2+
0.0013746 M
[Pb ] =
0.001715329 M
[sol.lead]=
Compare to
*Without considering other soluble lead iodide species,
we can say 2[Pb2+] = [I-], and let x =[Pb2+]
Ksp = x(2x)2 = 4x3 , x = [Pb2+] = 0.001255 M
Figure 6-2 in text
log[soluble Pb]
0
-3
-2
-1
-1
0
1
-2
-3
-4
-5
log[I-]
Example 2:
Place lead iodide in water and add sodium iodide, such that [I-] = 0.1 M
Write the pertinent reactions.
PbI2(s) ↔ Pb2+ +2IKsp = 7.9*10-9 = [Pb2+][I-]2
Pb2+ + I- ↔ PbI+(aq)
K1 = 100 =[PbI+(aq)]/[Pb2+][I-]
2+
Pb + 2I ↔ PbI2(aq)
2 = 1400 = [PbI2(aq)]/[Pb2+][I-]2
Pb2+ + 3I- ↔ PbI3-(aq)
3 = 8300 = [PbI3-(aq)]/[Pb2+][I-]3
Pb2+ + 4I- ↔ PbI42-(aq)
4 = 30000 = [PbI42-(aq)]/[Pb2+][I-]4
Ksp = x(0.1)2 = 7.9*10-9
So,
x = 7.9*10-7 M
Soluble lead = 2.86×10-5 M
common ion effect!
Example 3:
Place lead iodide in water and add sodium iodide, such that [Na+] = 1.0 M
PbI2(s) ↔ Pb2+ +2IKsp = 7.9*10-9 = [Pb2+][I-]2
Pb2+ + I- ↔ PbI+(aq)
K1 = 100 =[PbI+(aq)]/[Pb2+][I-]
2+
Pb + 2I ↔ PbI2(aq)
2 = 1400 = [PbI2(aq)]/[Pb2+][I-]2
Pb2+ + 3I- ↔ PbI3-(aq)
3 = 8300 = [PbI3-(aq)]/[Pb2+][I-]3
Pb2+ + 4I- ↔ PbI42-(aq)
4 = 30000 = [PbI42-(aq)]/[Pb2+][I-]4
Let x = [Pb2+],
Then
Soluble lead = x + K1x(1) + 2x(1)2 + 3x(1)3 + 4x(1)4
Soluble lead = x + 100x + 1400x + 8300x + 30000x
≈ 39801 x
[Pb2+] = x = Ksp/[I-]2
assume that very little [I-] reacts, so that [I-] ≈ 1.0 M
x = 7.9*10-9/(1)2 = 7.9*10-9M
Soluble lead = 0.000314 M
Complex ion formation dominates!
Figure 6-2 in text
log[soluble Pb]
0
-3
-2
-1
-1
0
1
-2
-3
-4
-5
log[I-]
Chapter 10 Monoprotic Acids and Bases
Bronsted and Lowry definitions
Acids are proton donars
Bases are proton acceptors
HCl(g) + NH3(g)  NH4+Cl-(s)
acid
base
salt
CH3OOH + CH3NH2
↔
Acetic acid methyl amine
CH3OO- +
CH3NH3+
acetate ion
methyl ammonium
ion
acid
base
conjugate base
conjugate acid
For aqueous solutions,
Acids react with water to yield H3O+
Bases react with water to yield OHacid
CH3OOH + H2O
↔
CH3OO- +
H3O+
Acetic acid
water
acetate ion
hydronium ion
acid
base
conjugate base
conjugate acid
conjugate pairs
CH3OOH
CH3OO-
base
CH3NH2 +
methyl amine
base
H2 O
water
acid
↔
CH3NH3+
+
methyl ammonium ion
conjugate acid
conjugate pairs
CH3NH2
CH3NH3+
Autoprotolysis of water
H2O + H2O ↔ H3O+ + OHEquilibrium constant is Kw
Kw = [H3O+][OH-] = 1.01*10-14 @ 25 C
For pure water (what is pure water?)
[H3O+] = [OH-] = 1.00*10-7 M
OHhydroxide ion
conjugate base
pH – a measure of the acidity of a solution
pH = -log [H3O+] or –log AH+ (ch. 8)
below 7 is acidic
above 7 is neutral
Strong acids and bases – complete dissociation in water
Example of strong acid
What is the pH of a 0.2 M HCl solution
HCl is a strong acid
HCl + H2O → Cl- + H3O+
The equilibrium constant for this reaction is large. There is virtually no
HCl species remaining in solution. Therefore,
[H3O+] = 0.2 M and pH = -log (0.2) = 0.7
0.02 M solution → 1.7
sig. fig. rule: the number of significant figures in the
concentration is equal
to the number of significant figures after
the decimal point in the pH.
Example of strong base
What is the pH of a 0.020 M Ca(OH)2 solution?
Ca(OH)2 is a strong base.
Ca(OH)2 → Ca2+ + 2OHThe equilibrium constant for this reaction is large. There is virtually no
Ca(OH)2(s) remains. Therefore,
[OH-] = 0.040 M, [H3O+] = 2.5*10-13 M, and pH = -log (0.040) = 12.60
sig. fig. rule: the number of significant figures in the
concentration is equal
to the number of significant figures after
the decimal point in the pH.
However, in these examples we have made an assumption. The assumption that
has implicitly been made is that the [H+] from the strong acid (or [OH-] from the
strong base) is >>>>> the [H+] (or the [OH-]) from water hydrolysis. Certainly
these assumptions have been true!!!
In the strong acid example,
[H+]sa = 0.1 M
Kw = [OH-][H3O+] = 1.01*10-14
[OH-] = 1.01*10-13 M
[H3O+]water = [H+]water = 1.01*10-13 M
but what happens if this assumption breaks down?
Example: Problem 10-3
Calculate the pH of a 5.0*10-8 M HClO4 solution
+
H3O and
HClO4 is a strong acid, and thus completely dissociates to give
ClO4- ions.
Sources of H3O+
HClO4 + H2O → H3O+ + ClO42H2O  H3O+ + OHpertinent equations;
[H3O+] = [H3O+]HClO4 + [H3O+]H2O
[OH-] = [H3O+]H2O
1.01*10-14 = Kw = [H3O+] [OH-]
[H3O+]HClO4 = 5.0*10-8 M
let x = [H3O+]H2O
substitutions;
Kw = x(5.0*10-8 + x) = 1.01*10-14
assumption
Assumption 1: Is [H3O+]H2O <<< [H3O+]HA
If at first we make the assumption, we soon see that …
[H3O+]HA = 5.0*10-8 M and [H3O+]H2O = 2.0*10-7 M, and the
is not valid!!!!
Back to the original expression, x(5.0*10-8 + x) = 1.01*10-14
This is a quadratic. Using excel, successive approximations
gives,
x = 7.86*10-8 M
[H3O+] = 5.0*10-8 + x = 1.286*10-7 M,
pH = -log(1.286*10-7) = 6.89
This result makes sense. The pH of a dilute strong acid ought to be
slightly acidic.
Weak Acids and Bases
Treatment of Weak acids – HA
HA + H2O ↔ A- + H3O+
Equilibrium constant is Ka
Ka = [A-][H3O+]/[HA]
The autoprotolysis reaction must also be satisfied
2H2O ↔ OH- + H3O+
Kw = [OH-][H3O+] = 1.01*10-14
Ka = [A-]([H3O+]HA + [H3O+]H2O)/([HA]0 – [A-])
[H3O+]HA
[H3O+]H2O
[HA]0
Define:
Assumptions that can often be made:
(1)
[H3O+]H2O <<< [H3O+]HA, therefore HA is the only source
of H3O+, and [A-] = [H3O+]
(2)
If Ka is small, [A-] <<< [HA], and [HA] = [HA]0
Example of a weak acid problem:
Calculate the pH of a 0.10 M solution of acetic acid.
Ka = 1.75*10-5
CH3OOH
+
H2O
↔
CH3OO-
+
H3O+
Ka
= 1.75*10-5 = [CH3OO-] [H3O+] / [CH3OOH]
Let x = [CH3OO-]
Ka
= x(x + [H3O+]H2O) / (0.10 – x)
Make assumptions:
(1)
[H3O+]H2O <<< [H3O+]HA
(2)
[HA] = [HA]0
The Ka expression reduces to
Ka
= x2 / 0.10 = 1.75*10-5
x = 0.001322876 M = [CH3OO-] = [H3O+]
pH = -log(0.001322876) = 2.88 (2 sig fig after the decimal pt.)
Are the assumptions valid?
Assumption 1:
Is [H3O+]H2O <<< [H3O+]HA?
[H3O+] = 0.00132 M, so [OH-] = Kw/[H3O+] = 7.65*10-12
M
The only source of the [OH-] is from autoprotolysis of
therefore, [H3O+]H2O = [OH-] = 7.65*10-12 M
Assumption 1 is valid
water,
Assumption 2:
Is [HA] >> [A-]?
If [A-]/[HA]0 < 0.05, we will call the assumption valid
In other words; the extent of dissociation is less than 5
%
[A-]/[HA]0 = (.00132/0.1) = 0.0132, assumption is valid
Treatment of Weak bases, B
B + H2O ↔ BH+ + OHEquilibrium constant is Kb
Kb = [BH+][OH-]/[B]
The autoprotolysis reaction must also be satisfied
2H2O ↔ OH- + H3O+
Kw = [OH-][H3O+] = 1.01*10-14
Kb = [BH+]([OH-]B + [OH-]H2O)/([B]0 – [BH+])
Define:
[OH-]HA
[OH-]H2O
[B]0
Assumptions that can often be made:
(3)
[OH-]H2O <<< [OH+]B, therefore B is the only source of
OH-, and [BH+] = [OH-]
(4)
If Kb is small, [BH+] <<< [B], and [B] = [B]0
Example of a weak base
Calculate the pH of a 0.10 M solution of sodium acetate.
Sodium acetate is a soluble salt
NaAc → Na+ + Ac-
CH3OO- + H2O
↔
CH3OOH + OH-
Kb(acetic acid) is the equilibrium constant for this reaction
Kb(acetic acid) = Kw/Ka(acetic acid) = 1.01*10-14/1.75*10-5 =
5.77*10-10
Kb = 5.77*10-10 = [CH3OOH] [OH-] / [CH3OO-]
Let x = [CH3OOH]
Kb = x(x + [OH-]H2O) / (0.10 – x)
Make assumptions:
(1)
[OH-]H2O <<< [OH-]A(2)
[A-] = [A-]0
The Kb expression reduces to
Kb = x2 / 0.10 = 5.77*10-10
x = 7.596991886*10-6 M = [CH3OOH] = [OH-]
[H3O+] = Kw/[OH-] = 1.01*10-14/7.596991886*10-6 =
1.3294735*10-9 M
pH = -log(1.3294735*10-9) = 8.88 (2 sig fig after the decimal pt.)
Are the assumptions valid?
Assumption 1:
Is [OH-]H2O <<< [OH-]HA?
[OH-] = 7.60*10-6 M
therefore, [OH-]H2O = [H3O+] = 1.3*10-9 M
If [OH-]H2O / [OH-] < 0.05, we will call
the assumption valid
[OH-]H2O / [OH-] = 1.3*10-9/7.60*10-6 =0.0002
Assumption 1 is valid
Assumption 2:
Is [B] >> [BH+]?
If [BH+]/[B]0 < 0.05, we will call the assumption valid
In other words; the extent of dissociation is less than 5
%
[BH+]/[B]0 = (7.6*10-6/0.1) = 7.6*10-5, assumption is
valid
For weak acids and bases, if assumption 1 fails it is a little more tricky.
Some tricks that we will pick up from Chapter 9 are useful in such a case.
What if assumption 2 is not valid?
What is the pH of a 5.0 mM solution of ammonia?
Ammonia is a weak base
Kb = Kw/Ka(NH4+) = 1.01*10-14/5.70*10-10 = 1.77*10-5
CH3OO- + H2O ↔ CH3OOH + OHKb
= 1.77*10-5 = [NH4+] [OH-] / [NH3]
Let x = [NH4+]
Kb
= x(x + [OH-]H2O) / (0.005 – x)
Make assumptions:
(1)
[OH-]H2O <<< [OH-]NH3
(2)
[NH3] = [NH3]0
The Kb expression reduces to
Kb
= x2 / 0.0050 = 1.77*10-5
x = 2.97489*10-4 M = [CH3OOH] = [OH-]
+
[H3O ] = Kw/[OH-] = 1.01*10-14/2.97489*10-4= 3.3950779*10-11
M
pH = -log(3.3950779*10-11) = 10.47 (2 sig fig after the decimal
pt.)
Are the assumptions valid?
Assumption 1:
Is [OH-]H2O <<< [OH-]HA?
[OH-] = 2.97*10-4 M
therefore, [OH-]H2O = [H3O+] = 3.40*10-11 M
If [OH-]H2O / [OH-] < 0.05, we will call
the assumption valid
[OH-]H2O / [OH-] = 3.40*10-11/2.970*10-4 <<<< .05
Assumption 1 is valid
Assumption 2:
Is [B] >> [BH+]?
If [BH+]/[B]0 < 0.05, we will call the assumption valid
In other words; the extent of dissociation is less than 5
%
[BH+]/[B]0 = (2.970*10-4 /0.005) = 0.06,
assumption is not valid!!!!
Rewrite expression without assumption 2!!
Kb
= x2 / (0.0050-x) = 1.77*10-5
This is a quadradic
x2 + 1.77*10-5x - (0.005)(1.77*10-5)
You can use Excel and solve by successive approximation
x = 2.8877 *10-4 M instead of (2.970*10-4)
[H3O+] = 1.01*10-14/ 2.8877 *10-4 = 3.4976*10-11
pH = 10.46
Another example:
Calculate the pH of a 0.010 M solution of chlorous acid.
Ka = 1.12*10-2
HOClO + H2O ↔ OClO- + H3O+
Ka
= 1.12*10-2 = [OClO-] [H3O+] / [HOClO]
Let x = [CH3OO-]
Ka
= x(x + [H3O+]H2O) / (0.010 – x)
Make assumptions:
(1)
[H3O+]H2O <<< [H3O+]HA
(2)
[HA] = [HA]0
The Ka expression reduces to
Ka
= x2 / 0.010 = 1.12*10-2
x = 0.0105838 M = [CH3OO-] = [H3O+]
Are the assumptions valid?
water,
Assumption 1:
Is [H3O+]H2O <<< [H3O+]HA?
[H3O+] = 0.0106 M, so [OH-] = Kw/[H3O+] = 9*10-13 M
The only source of the [OH-] is from autoprotolysis of
therefore, [H3O+]H2O = [OH-] = 7.65*10-12 M
Assumption 1 is valid
Assumption 2:
Is [HA] >> [A-]?
Clearly not!!!
Using excel solve the full expression for x
1.12*10-2 = x(x + [H3O+]H2O) / (0.010 – x)
x = 0.006181 M
pH = 2.21
Buffers
A buffer is a solution that is resistant to pH changes. It usually consist
of a mixture of an acid and its conjugate base in roughly equal proportions
Let’s look at weak acid, HA (Ka = 1.00*10-4)
HA + H2O ↔ A- + H3O+
Ka = [A-] [H3O+] / [HA] = [H3O+] {[A-] / [HA] }
Take the log of both sides
Log Ka = log [H3O+] + log {[A-] / [HA]}
-pKa = -pH + log {[A-] / [HA]}
pH = pKa + log {[A-] / [HA]} Henderson-Hasselbach
equation
Implication:
its
relatively
can rewrite
For a solution containing an equal mixture of an acid and
conjugate base, pH = pKa, and this result is
independent of pH!!!
Since [HA] and [A-] are components of the same solution, we
the Henderson-Hasselbach equation as follows:
pH = pKa + log (mol A-] / mol HA)
As we shall see, it is often more convenient in this form.
How does a buffer work?
Add 1.0 ml of concentrated HCl to 1.0 L of water. What is the
pH?
Concentrated HCl is 12.1 M
1 ml of concentrated HCl = 12.1 mmol
[H3O+] = (12.1 mmol)/1001 ml = 0.012 M
pH = -log(.012) = 1.92
0.50 mol
(NaA).
Now add 1.0 ml of concentrated HCl to 1.0 L solution containing
of acetic acid (HA) and 0.50 mol of sodium acetate
Ka(acetic acid) = 1.75*10-5
pKa = 4.757
pH before the addition of the HCl
pH = 4.757 + log(mol A- / mol HA) = 4.757 + log
(0.50/0.50) =
4.76
strong
pH after the addition of the HCl
The HCl reacts rapidly and completely with the A- (it is a
acid)
HCl + A- → Cl- + HA
Initial conditions:
HCl: 12.1 mmol
A-:
500 mmol
HA: 500 mmol
Final conditions:
HCl: 0.00 mmol
A-:
(500-12.1) = 487.9 mmol
HA: (500+12.1) = 512.1 mmol
(487.9/512.1) =
pH = 4.757 + log(mol A- / mol HA) = 4.757 + log
4.74
Try adding 1.0 ml of concentrated HCl to 1.0 L solution containing
0.050 mol of acetic acid (HA) and 0.050 mol of sodium acetate (NaA).
Initial conditions:
HCl: 12.1 mmol
A-:
50 mmol
HA:
50 mmol
Final conditions:
HCl: 0.00 mmol
A-:
(50-12.1) = 37.9 mmol
HA: (50+12.1) = 62.1 mmol
pH = 4.757 + log(mol A- / mol HA) = 4.757 + log (37.9/62.1) =
4.54
Try adding 1.0 ml of concentrated HCl to 1.0 L solution containing
0.0050 mol of acetic acid (HA) and 0.0050 mol of sodium acetate (NaA).
Initial conditions:
HCl: 12.1 mmol
A-:
5.0 mmol
HA: 5.0 mmol
Final conditions:
HCl: 7.1 mmol
A-:
(5.0-5.0) = 0 mmol
HA: (5.0+5.0) = 10 mmol
pH = -log(.0071) = 2.2, the capacity of the buffer solution was
exceeded
Why doesn’t HA will contribute somewhat to the [H3O+]?
Ka = [H+][A-]/[HA] = (0.1+x)x/(0.1-x) = 1.75*10-5 , x =
1.75*10-5
Buffer capacity (b) = dCb/dpH = -dCa/dpH
Depends upon
1)
ratio of A-/HA, closer to 1, the greater the buff.
Capacity
2)
the absolute concentration of A- and HA
Lets do one more!!!
Try adding 1.0 g of concentrated NaOH to 1.0 L solution containing
0.050 mol of acetic acid (HA) and 0.050 mol of sodium acetate (NaA).
NaOH + HA  Na+ + H2O + AThe NaOH reacts rapidly and completely with the HA (it is a
strong base)
Initial conditions:
NaOH: (1.0 g)/(40.0 g/mol) = 25 mmol
A-:
50 mmol
HA: 50 mmol
Final conditions:
NaOH: 0.00 mmol
A-:
(50+25) = 75 mmol
HA: (50-25) = 25 mmol
pH = 4.757 + log(mol A- / mol HA) = 4.757 + log (75/25) = 5.23
Dilute buffer problem
What is the pH of a solution prepared by mixing 0.0100 mol HA
(with Ka
= 1.00*10-2) and 0.0100 mol of A- and diluting
with water to 1.00 L?
can occur
Because HA is a moderately strong acid, the following reaction
to an appreciable extent.
HA + H2O ↔ A- + H3O+
Ka = [H3O+][A-]/[HA]
pH = pKa + log {[A-] / [HA]}
[A-] = FA- + [H3O+]
[HA] = FHA - [H3O+]
FA- = FHA = 0.0100 M
Let [H3O+] = x
Ka = x(.0100+x)/(.0100-x)
1.00*10-2 = x(.0100+x)/(.0100-x)
Excel, successive approximations → x = 0.00414 M
[A-] = FA- + [H3O+] = 0.0100 + 0.00414 =
0.01414 M
[HA] = FHA - [H3O+] = 0.0100 – 0.00414 =
0.00586 M
pH = pKa + log {[A-] / [HA]} = 2.000 +
log(0.01414/0.00586) = 2.383
or alternatively, since you already have [H3O+] = x
pH = -log x = 2.383