Approximation for pH calculation with diluted acids in weak-strong titration curve

Question

I was wondering how one could calculate small pH changes in titrations with dilute acids.

As we know, usually we approximate the pH by saying that the conversion is quantitative. For example, when we titrate i.e 100ml 0.1M sodium acetate solution with hydrchloric acid, the start point would be just pOH=0.5(pKb-log(c)), with pKb=9.25 we get 5.125. So pH=14-5.125=8.875.

If we add 25ml of 0.1M HCl we say that HCl consumes acetate directly:

CH₃COO⁻ + HCl → CH₃COOH + Cl⁻

So n(HCl)=0.0025mol consumes equal amounts of acetate and forms 0.0025mol of acetic acid. 0.0075 moles of acetate remain. We get 125ml of solution, so c(Ac-)=0.06M and c(HAc)=0.02M, so Hendersson-Hasselbalch gives us pH=pKs+log(0.06M/0.02M)=4.75+log(3)=5.227.

That in itself makes sense, but it doesn't if we, for example, only titrate with 10⁻⁶M HCl? If we added 0.5L of that, according to this approximation we would have n(HAc)=0.5*10⁻⁶mol and n(Ac-)=0.0099995mol both in 0.6L of solution, and subsequently a higher pH value than if we just let the acetate dissociate in pure water. So how can we replace this approximation in this, admittedly, extreme case so that the titration curve can predict the real result?

Answer 1

In such cases, you cannot afford to involve implicit simplifications.

In this particular case, the equilibrium ratio $\frac {[A-]}{[\ce{HA}]}$ is not the same as the nominal $\frac {[A-]}{[\ce{HA}]}$ ratio, but reaching acido-basic equilibrium is to be considered. One has to count also with water auto-ionization equilibrium.

If the added $\ce{HCl}$ converted the full equivalent amount of acetate to acetic acid, $\ce{pH}$ would not change. A small portion of acetate will not convert to acid and the respective portion of remaining $\ce{H+}$ will add itself to acidity of the solution, all together honoring the acidity constant.

Full analysis, involving mass+charge balance/conservation and ongoing equilibria is to be evaluated and only those simplifications that are justified can be accepted.(1)

• Charge balance: $\ce{[H+]} + \ce{[Na+]}= \ce{[OH-]} + \ce{[A-]} +\ce{[Cl-]}$
• Weak acid mass balance : $c = \ce{[HA]} + \ce{[A-]}$
• Weak acid mass balance 2 : $c = \ce{[Na+]}$
• Acidity constant: $K_\mathrm{a} = \frac{\ce{[H+][A-]}}{\ce{[HA]}}$
• Water auto-ionization: $K_\mathrm{w} = \ce{[H+][OH-]}$

By solving 5 equations with 5 variables, we get $\ce{[H+]}$. Justified simplifications are always welcome.

Substituting for $\ce{[OH-]}$ and $\ce{[Na+]}$:

• Charge balance: $\ce{[H+]} + c = \frac {K_\mathrm{w}}{\ce{[H+]}} + \ce{[A-]} +\ce{[Cl-]}$
• Weak acid mass balance : $c = \ce{[HA]} + \ce{[A-]}$
• Acidity constant: $K_\mathrm{a} = \frac{\ce{[H+][A-]}}{\ce{[HA]}}$

Substitution with last 2 equations:

• Charge balance: $\ce{[H+]} + c = \frac {K_\mathrm{w}}{\ce{[H+]}} + \ce{[A-]} +\ce{[Cl-]}$
• Weak acid mass balance : $c = \frac{\ce{[H+][A-]}}{K_\mathrm{a}} + \ce{[A-]} = \ce{[A-]} \cdot (\frac{\ce{[H+]}}{K_\mathrm{a}} + 1) \implies \ce{[A-]} = \frac {c}{\frac{\ce{[H+]}}{K_\mathrm{a}}+1} = \frac {c \cdot K_\mathrm{a}}{\ce{[H+] + K_\mathrm{a}}}$

and finally:

• Charge balance: $\ce{[H+]} + c = \frac {K_\mathrm{w}}{\ce{[H+]}} + \frac {c \cdot K_\mathrm{a}}{\ce{[H+] + K_\mathrm{a}}} +\ce{[Cl-]}$

Multiplied by denominators:

$\left(\ce{[H+]} + c \right)\ce{[H+]} \cdot \left(\ce{[H+]} + K_\mathrm{a}\right) = K_\mathrm{w}\left(\ce{[H+]} + K_\mathrm{a}\right) + c \cdot K_\mathrm{a} \cdot \ce{[H+]} +\ce{[Cl-]}\ce{[H+]} \cdot \left(\ce{[H+]} + K_\mathrm{a}\right)$

Separated by the power order: ${\ce{[H+]}}^{3} + {\ce{[H+]}}^{2} \cdot (c + K_\mathrm{a} - \ce{[Cl-]}) - \ce{[H+]} \cdot ( K_\mathrm{w} + \ce{[Cl-]} \cdot K_\mathrm{a}) - K_\mathrm{w} \cdot K_\mathrm{a} = 0$

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(1)- The classical related case is $\mathrm{pH} = \frac 12 ( \mathrm{p}K_\mathrm{a} - \log{c} )$, implying $c \gg [\ce{H+}] \gg [\ce{OH-}]$