Transcript Chapter 13 Advanced Topics in Equilibrium

Chapter 13
Advanced Topics in Equilibrium
Overview
13-1 General Approach to Acid-Base
Systems
13-2 Activity Coefficients
13-3 Dependence of Solubility of pH
13-4 Analyzing Acid-Base Titrations with
Difference Plots
Advanced Topics in Equilibrium
• Calcite (CaCO3) is not very soluble in
neutral or basic solution(Ksp = 4.5 x 10-9).
• Calcite dissolves in acid by virtue of two
coupled equilibria, in which reactions have
a species in common.
2
CaCO 3 ( s )  Ca  CO
CO
23

 H  HCO
3
23
13-1: General Approach to Acid-Base
Systems
Example: Determine the pH and concentration of all
species in a solution of 20.0 mmol sodium tartrate (Na+HT), 15.0 mmol pyridinium chloride (PyH+Cl-), and 10.0 mmol
KOH, in a volume of 1.00 L.
The reactions and Ka values are:
The charge balance is:
13-1: General Approach to Acid-Base
Systems
The mass balances are:
[Na+] = 0.020 0 M [K+] = 0.0100 M [Cl-] = 0.015 0 M
[H2T] + [HT-] + [T2-] = 0.020 0 M [PyH+] + [Py] = 0.015 0 M
 10 independent equations = 10 species
Step 1 Write a fractional composition equation for each acid or
base that appears in the charge balance. Designate the formal
concentrations as FH2T = 0.0200 0M and FPyH+ = 0.015 0 M.
13-1: General Approach to Acid-Base
Systems
Step 2 Substitute the fractional composition expressions
into the charge balance and enter known values for [Na+],
[K+], and [Cl-]. Also, write [OH-] = Kw/[H+].
Step 3 Use a spreadsheet to solve for [H+].
Guess a value for pH in cell H13 and use Excel Solver to
vary pH until it satisfies the charge balance – the sum of
charges in cell E15 is 0.
13-1: General Approach to Acid-Base
Systems
13-2: Activity Coefficients
• The Davies equation for calculating activity coefficients
has no size parameter:
• Consider a primary standard buffer containing 0.025 0 m
KH2PO4 and 0.025 0 m Na2HPO4. Its pH (25°C) is 6.865
± 0.006. With m = 0 at 25°C, for H3PO4:
13-2: Activity Coefficients
• When m is not zero, activity coefficients are incorporated
into an effective equilibrium constant, K′, at a given ionic
strength:
• Activity coefficients for the ionic species are computed
using the Davies equation; neutral species g ∞ 1.00.
13-2: Activity Coefficients
• Recall the equations for water ionization:
H 2 O  H   OH - K w  [H  ]γ H  [OH - ]γ OH - 10 -13.995
• Find the pH of 0.025 0 m KH2PO4 plus 0.025 0 m Na2HPO4
including activity coefficients using the chemical reactions
13-19 through 13-21, the water ionization, and the mass
balance: [K+] = 0.025 0 m, [Na+] = 0.050 0 m, and FH3P =
0.050 0 m.
• The charge balance is:
13-2: Activity Coefficients
• Substitute expressions into the charge balance to obtain an
equation in which [H+] is the only variable.
• Use the fractional composition equations for H3PO4:
• In the spreadsheet – input values in the shaded cells and
guess a value for pH in cell H15.
• Use Excel Solver to vary the pH in cell H15 such that the
charge in cell E18 is nearly zero.
13-2: Activity Coefficients
13-3: Dependence of Solubility on pH:
Solubility of CaF2
• The mineral fluorite (CaF2), also called fluorspar is
converted to HF for the synthesis of refrigerants and
fluoropolymers.
13-3: Dependence of Solubility on pH
• The solubility of CaF2 is governed by Ksp for the salt,
hydrolysis of F- and Ca2+, and ion-pairing between Ca2+
and F-:
Charge balance:
Mass balance:
13-3: Dependence of Solubility on pH
• Figure 13-5 shows the spreadsheet for CaF2, in which
we begin with the estimates pCa2+ = 4 and pH = 7 in cells
B8 and B9.
13-3: Dependence of Solubility on pH
• Cells D17:D21 compute the effective equilibrium
constants K′. using the activity coefficients and cells
C10:C14 compute concentrations from these equilibrium
constant expressions:
13-3: Dependence of Solubility on pH
• Activity coefficients are
computed in cells E8:F14
from the Davies equation
using the ionic strength
computed in cell B5.
• Use Excel Solver to vary
pCa2+ and pH to minimize
cell J19.
• Then alternately hold pCa2+
constant while solving for pH
and vice versa, repeating the
cycle until J19 stops
decreasing.
13-3: Dependence of Solubility on pH:
Solubility of Barium Oxalate
• Consider the dissolution of Ba(C2O4) whose
anion is dibasic and whose cation is a weak
acid.
Charge balance:
Mass balance:
13-3: Dependence of Solubility on pH
• We have 8 unknowns and 8 independent equations (don’t
forget [OH-] = Kw/[H+]).
• We deal with ion paring by adding reactions 13-44 and 13-48
to find:
• So, [Ba(C2O4)(aq) = 10-4.54 M as long as undissolved
Ba(C2O4)(s) is present.
• Abbreviating oxalic acid as H2Ox we write the fractional
composition equations:
13-3: Dependence of Solubility on pH
• Ba2+ and BaOH+ are a conjugate acid-base pair with Ba2+
behaving as a monoprotic acid HA, and BaOH+ as the
conjugate base A-.
Fixing the pH with a buffer and substituting FBa= FH2Ox:
13-3: Dependence of Solubility on pH
• Figure 13-9 shows the
spreadsheet used to
find the pH in cell A11
necessary to make the
net charge in cell H23
equal to 0.
13-3: Dependence of Solubility on pH
• The solubility of barium
oxalate is steady in the
middle pH range.
• Solubility increases below
pH 5 because C2O42reacts with H+ to make
HC2O4-.
• Solubility decreases above
pH 13 because Ba2+ reacts
with OH- to make BaOH+.
13-4: Analyzing Acid-Base Titrations
with Difference Plots
• A difference plot (Bjerrum plot) can be used to extract
metal-ligand formation constants or acid dissociation
constants from titration data obtained with electrodes.
• Let’s apply a difference plot to an acid-base titration curve.
• For a diprotic acid (H2A), the mean fraction of proton
bound ranges from 0-2:
• It is measured by a titration beginning with a mixture of A
mmol of H2A and C mmol of HCl in V0 mL.
• HCl is added to increase the degree of protonation of H2A.
• After titrating with v mL of Cb (mol/L) standard NaOH,
mmol Na+ = Cbv.
• The net charge balance is:
13-4: Analyzing Acid-Base Titrations
with Difference Plots
• The denominator of Equation 13-58 is:
FH2A = [H2A] + [HA-] + [A2-]
• the numerator can be written as:
2FH2A – [HA-] – 2[A2-]
• therefore:
• From Equation 13-59, we can write:
-[HA-] – 2[A2-] = [OH-] + [Cl-]HCl – [H+] – [Na+]
• Substituting this into Equation 13-60’s numerator:
nH 
2FH 2 A  [OH - ]  [Cl - ]HCl - [H  ] - [Na  ]
FH 2 A
[OH - ]  [Cl - ]HCl - [H  ] - [Na  ]
2
FH 2 A
13-4: Analyzing Acid-Base Titrations
with Difference Plots
• For HnA, the mean fraction of bound protons turns out to
be:
• Each term on the right side of the equation is known during
the titration.
• [H+] and [OH-] are measured with a pH electrode.
• Using Equation 13-25 and pH = -log([H+]gH+, substituting
into Equation 13-61, we get the measured fraction of bound
protons:
13-4: Analyzing Acid-Base Titrations
with Difference Plots
• In acid-base titrations, a difference plot is a graph of the
mean fraction of protons bound to an acid versus pH.
• For complex formation, the difference plot gives the mean
number of ligands bound to a metal versus pL.
• Equation 13-62 gives the measured fraction of bound
protons.
• For a diprotic acid, the theoretical mean fraction of bound
protons is:
• Figure 13-11 shows experimental data
for a titration of the amino acid glycine.
13-4: Analyzing Acid-Base Titrations
with Difference Plots
13-4: Using Excel to Optimize More
Than One Parameter
• In Figure 13-11, we want values in cells B9, B10, and B11
that minimize the sum of the square of the residuals in cell
B12.
• When using Solver to optimize several parameters at one
time, try different starting values to see if the same solution is
reached.