Acid Chemistry

You probably noticed that on the pH scale, different chemicals had different pH’s. How can you determine what the pH of an acid is?

Well, we’ve already actually calculated that. Remember when we looked at the pH of pure water? We were treating water as an acid. Let’s instead try to calculate the pH of a solution of $CH_3COOH$, acetic acid.

Sample Problem 6.3.1
What is the pH of a 3M solution of Acetic Acid?
$CH_3COOH\leftrightarrow CH_3COO^-+H^+$ $K=10^{-4.76}$

Notice that our equilibrium constant for this problem was $10^{-4.76}$. For acids, equilibrium constants are typically expressed as a pKa. Ka here represents the equilibrium constant for the acid dissociation constant.

$pK_a=-\log[K_a]$

$K_a$ is the equilibrium constant for $HA\leftrightarrow H^++A^-$, where HA is a generic acid.

For a resolved conjugate pair, pH=pKₐ when its conjugate-base/acid activity ratio is one. In the usual ideal buffer approximation, this means equal concentrations. Half-equivalence in a titration of a weak monoprotic acid with strong base is one way to prepare that ratio.

When a solution is at the pKa of a constituent acid, that acid is equally present as $HA$ as it is $A^-$.

In order to prove this, we start with the equilibrium expression.

$K = \frac{[H^+][A^-]}{[HA]}$

Now let’s take the -log for each side:

$-\log K_a=-\log\frac{[H^+][A^-]}{[HA]}$

$-\log K_a=-\log[H^+]-\log\frac{[A^-]}{[HA]}$

$pK_a=pH-\log\frac{[A^-]}{[HA]}$

$pH = pK_a+\log\frac{[A^-]}{[HA]}$

Now, if the acid is 50% deprotonated, then $\frac{[A^-]}{[HA]}=1$, and:

$pH = pK_a+\log(1)$

$pH = pK_a$

That equation we reached before plugging in our 50% protonation assumption is called the Henderson-Hasselbalch equation, and it is used to relate pH to the dissociation of an acid.

Aside from using the Henderson-Hasselbalch equation to solve for the pH, you can also use it to determine the concentration of an acid.

Sample Problem 6.3.2
A solution of Hydrofluoric Acid ($pK_a=3.17$) has a pH of 2. What is the concentration of the acid?

One final use of the Henderson-Hasselbalch Equation is to predict whether the acid or conjugate base is dominant in a solution.

Example Problem 6.3.3
Is Ammonia or Ammonium more predominant in a solution if the pH is 10? $pK_a=9.25$ for Ammonium.

Henderson–Hasselbalch can describe a particular conjugate pair of a polyprotic acid when that pair dominates and the approximations hold. It is not restricted to molecules with only one acidic proton. Full quantitative polyprotic distributions are retained enrichment beyond this CED’s required AP calculations.

To start off with, let’s go through what happens with a diprotic acid. We’ll take Sulfuric Acid as our diprotic acid.

Polyprotic acids have their $pK_a\text{'s}$ given with the order of dissociation. For sulfuric acid, they look like the following:

$H_2SO_4\leftrightarrow H^++HSO_4^-$ $pK_{a1}=-6.62$

$HSO_4^-\leftrightarrow H^++SO_4^{2-}$ $pK_{a2}=1.99$

Alright, so let’s start off by writing up the equilibrium equation for the initial deprotonation.

$K_1=10^{6.62}=\frac{[H^+][HSO_4^-]}{[H_2SO_4]}$

Now let’s set up the second equilibrium

$K_2 = 10^{-1.99}=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

Let’s set up an ICE table to keep track of all these components.

$\begin{matrix} \bold{Condition}&\bold{[H_2SO_4]}&\bold{[HSO_4^-]}&\bold{[SO_4^{2-}]}&\bold{[H^+]}\\ Initial&X&0&0&0\\ Change&-x&+x-y&+y&+x+y\\ Final&X-x&x-y&y&x+y \end{matrix}$

For this ICE table, we have to keep track of the progression of two reactions. The extent of the initial deprotonation is $x$, and the second is tracked with $y$. For this ICE table, we assumed there was only $H_2SO_4$ put into the solution, and that the Hydrogen Ions generated are significant enough to disregard the initial $10^{-7}M$.

Let’s put these values into the equilibrium equations.

$K_1=\frac{(x+y)(x-y)}{X-x}$

$K_2=\frac{(x+y)(y)}{x-y}$

Now we have 2 equations and three unknowns. This means that if we knew the initial concentration of $H_2SO_4$, or if we had another equation (say, by knowing $pH$, maybe?), we would be able to solve this problem.

Reviewed ideal 25 °C model: a phosphoric-acid-only solution has pH=1.20. Use pKₐ₁=2.12, pKₐ₂=7.21 and pKₐ₃=12.32, K_w=10⁻¹⁴ and no additional ions. Find total phosphate-family concentration C_T. Activities matter in a real concentrated solution.

Let h=[H⁺]=10⁻¹·²=0.06309573445 M and c₀,c₁,c₂,c₃ denote H₃PO₄,H₂PO₄⁻,HPO₄²⁻,PO₄³⁻. The successive equilibria give c₁=K₁c₀/h, c₂=K₁K₂c₀/h² and c₃=K₁K₂K₃c₀/h³.
Mass balance: C_T=c₀+c₁+c₂+c₃. Charge balance: h−K_w/h=c₁+2c₂+3c₃. These equations avoid treating later proton releases as independent extra acid amounts.
Substitute the equilibrium ratios into charge balance. c₀=(h−K_w/h)/(K₁/h+2K₁K₂/h²+3K₁K₂K₃/h³)=0.524806435 M.
The species concentrations are c₁=0.0630956111 M, c₂=6.16593797×10⁻⁸ M and c₃=4.67734227×10⁻¹⁹ M. The very small c₂ and c₃ reflect the correct second and third dissociation constants at this low pH.
Adding the four species gives C_T=0.587902107 M≈0.588 M. Check: c₁+2c₂+3c₃=0.063095734448 M=h−K_w/h. The original 0.5257 M answer and its intermediate substitutions are superseded.

We still haven’t defined what a strong or weak acid is, have we?

A strong acid transfers its proton to water essentially completely in the usual dilute aqueous model. Water levels sufficiently strong acids to hydronium. Acid strength depends on the solvent; percentage ionization of a weak acid also depends on concentration.

Some common examples of strong acids include $HI$, $HCl$, $HBr$, $HClO_4$, and $H_2SO_4$. Sulfuric acid is only strong for its first deprotonation.

H⁺(aq) is conventional shorthand for the hydrated proton, often written H₃O⁺. Treating H₃O⁺ and H⁺ as two independently measured acid species in water does not establish a universal percentage-dissociation cutoff.

Reviewed leveling-effect task: why do 0.0100 M HCl and 0.0100 M HBr have approximately the same pH in the ideal dilute aqueous model at 25 °C?

There is no universal 98.2% threshold separating strong and weak acids. For AP calculations, apply the stipulated dilute strong-acid model or solve the appropriate equilibrium; use activities when nonideality matters.