Theory behind the equiligraph
Theory behind the equiligraph
What makes this thing work?
The equiligraph is a graphic presentation of equations of state describing ionic equilibria. Most of us call these equations the dissociation constant, KD, or acid dissocation constant, Ka.
For example the dissociation constant of water:
Kw = (𝛾H * [H+] )* (𝛾OH * [OH-])
or for acetic acid, assuming all activity coefficients, 𝛾 = 1:
Ka = [H+]*[CH3COO-] / [CH3COOH]
The composition of a solution can be stated as a fraction of the total present in a given form to the the total concentration, Ct.
for a monoprotic acid;
[HB] = Ct * alpha0
[B] = Ct * alpha1
Ct = Ct*alpha0 + Ct * alpha1
1 = alpha0 + alpha1
where:
alpha0 = [HB] / Ct = [H] / ([H] + K1)
alpha1 = [B] / Ct = K1 / ([H] +K1)
for a diprotic acid:
1 = alpha0 + alpha1 + alpha2
alpha0 = [H2B] / Ct = [H]^2 / ([H]^2 + [H]*K1 + K1*K2)
alpha1 = [HB] / Ct = [H]*K1 / ([H]^2 + [H]*K1 + K1*K2)
alpha2 = [B] / Ct = K1*K2 / ([H]^2 + [H]*K1 + K1*K2)
Monoprotic example
1) Equilibria and constants
HA <--> H + B K1 = [H][B]/[HB]
H2O <--> H + OH Kw = [H][OH]
//*******************
2) Mass Balance
Ct = [HB] + [B]
//*******************
3) Charge Balance
[H] + [Na] = [OH] + [B]
//*******************
4) Couple the equations
autodissociation
small dissociation, large concentration assumptions, neglecting autodissocation of water:
[H] = [B] for pure acid dissolved in pure water, [B] >> sqrt(Kw), [B] << [HB]
[H] = sqrt ( K1 *[HB])
//*******************
Complete treatment, no assumptions
[HB] = Ct -[B] --> K1 = [H][B] / (Ct -[B])
K1*Ct - K1*[B] - [H]*[B] = 0
(K1+[H])*[B] = K1*Ct
[B] = (K1*Ct)/(K1+[H])
now couple with charge balance
[H] + [Na] = [OH] + [B] = kw/[H] + (K1*Ct)/(K1+[H])
(K1+[H])*[H]^2 + (K1+[H])*[Na] - (K1+[H])*Kw - K1*Ct = 0
[H]^3 + K1*[H]^2 + ([Na]-K)*[H] + ([Na] -Kw -Ct)*K1 =0
Note this is a cubic equation, so there are three critical points
//*******************
autoassociation
H20 <--> H + OH
- HB <--> H + + B
___________________________
B + H2O <--> BH + OH- Kb = [BH][OH]/ [B] assume [H2O] = 1
Kb = Kw / Ka
Small dissociation, large concentration, neglect contribution due to water
[BH] = [OH]
[OH] = sqrt(Kb*[B]) = sqrt ([B]*Kw/K1)
//*******************
5) buffer capacity
[Na] = [OH] + [B] -[H] = Kw/[H] + (K1*Ct)/(K1+Ct) - [H]
d[Na]/d[H] = -Kw/[H]^2 -(Ct*K1)/(K1+[H])^2 - 1
Now use chain rule
pH = -log10([H])
dpH/d[H] = -1/([H] *2.303)
d[Na]/pH = d[Na]/d[H] * d[H]/dpH = -1*(Kw/[H]^2 + Ct*(K1/(K1+[H])^2) +1)) * (-2.303*[H])
d[Na]/dpH = 2.303 *(Kw/[H] + (Ct*[H]*K1)/(Ct+[H])^2 + [H])
//*******************
alpha based on mass balance
Ct = [HB] + [B] = [HB] + Ct*K1/([H]+K1)
recast equilibrium to find [HB]
Ct = [HB] + [B] --> [B] = Ct -[HB]
K1 = [B][H]/[HB] --> (Ct -[HB])* [H] / [HB]
K1[HB] = [H]*Ct - [H]*[HB]
(K1 + [H])*[HB] = [H]*Ct/ (K1+[H])
mass balance now becomes
Ct = Ct*[H]/([H] + K1) + Ct*k1/([H] +k1)
1 = [H]/([H] +K1) + K1/ ([H] +k1) = alpha0 + alpha1
alpha0 =[H]/([H] +K1)
alpha1 = K1/ ([H] +k1)
These equations are originally cast in terms of the activity of each species. The activity is the true thermodynamic variable, however we often use the activity interchangeably with the molar concentration.
aX= 𝛾X* [X]
where:
aX = the activity of species X
𝛾X = the activity coefficient of species X
[X] is the molar concentration of species X
For simplicity, we will use concentration and activity interchangeably, except where a distinction is necessary.
We gain some simplicity when we constrain the concentration of related species using the mass balance equation.
Using a monoprotic acid, HB as an example:
CTotal = [HB] + [B]
1 = [HB] /CTotal + [B] / CTotal
The terms [HB] /CTotal and [B] / CTotal occur frequently, so we rename them as:
α0 = [HB] / CTotal
and
α1= [B] / CTotal xx
now the mass balance equation, now in terms of relative fractions, reduces to
1 = α0 + α1
and
[HB] = CTotal * α0
[B] = CTotal * α1
This conversion of concentration into relative fractions permits us to generalize the dissociation behavior on a log - log graph. Taking the log of the dissociation equations shown above:
log(Kw) = log(𝛾H * [H+] ) + log(𝛾OH * [OH-])
and
log(Ka ) = log( [H+]) + log([CH3COO-]) - log( [CH3COOH])
which are now linear functions
log(Ka ) = log( [H+]) + log(CTotal * α1 ) - log( CTotal * α0)
or
log(Ka ) = -pH + log(CTotal ) + log( α1 ) - log( CTotal ) - log(α0)
which reduces to:
- pKa = - pH + log( α1 ) - log(α0)
And this is identical to the more familiar form:
pH = log( [B] / [HB] ) + pKa
we now have log(α1 ) and log(α0) as linear functions of pH:
log(α0) = - pH + log( α1 ) + pKa
log( α1 )= pH + log(α0) - pKa
pKa is constant plots of log (α0) vs pH and plots of log(α1) vs pH have slopes of -1 and +1, respectively when the α1 ≃ 1 and α0 ≃ 1, respectively, which occurs when pH is far from pKa . This is shown in figure 1, which shows an equiligraph for acetic acid.
Keywords:
Figure 1. Equiligraph showing the composition of an acetic acid solution at pH = 4.1. The relative proportion of acid to conjugate base is fixed by the pH and to a good first order approximation is independent of concentration.
For a monoprotic acid or base, the rules for drawing the composition lines on a log-log plot, with the X axis defined as pH are:
1)When α0 or α1 are near unity, the line is horizontal and has a Y axis value = log(CTotal).
2)When α0 is near unity, α1 has a slope of +1.
3)When α1 is near unity, α0 has a slope of -1.
4)When 0.05 < α0 < 0.95 , then the lines are smooth curves connecting the composition lines and the two lines intersect at {pKa, log(CTotal ) - 0.3}
Notes:
log10(0.5) = -0.30, so [HB] = [B] = 0.5 * CTotal at Y = log(CTotal ) - 0.3.
log10(0.05) = 1.3 so:
α0 > 0.95 for pH < pKa - 0.65
and
α1 > 0.95 for pH > pKa + 0.65
Derivation of Analysis
The following derivation for a monoprotic acid is taken directly from the help files of the Titration 1.0 app.
The app’s help section also has derivations for diprotic, triprotic and tetraprotic acids.
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