12. Acid base equilibria (pH, Ka, pKa, Kw, pKw)

Two species that differ by a single H+ –> When an acid donates a proton, the conjugate base is what remains after the acid has lost a proton.

Proton [H+] donor

Proton [H+] acceptor

As a strong acid fully ionises, we use pH = -log(H+).

The concentration of the monoprotic acid is the same as the concentration of H+ ions

Ka acid constant Ka = [A-] [H+] / [HA] This is because a weak acid only partially ionises so the concentration of the acid wont equal H+ ions

The concentration of [H+] ions is the same as the conjugate base [A-], hence numerator of all Ka equations is written as [H^2] The concentration of the the intial acid [HA] is the same as the concentration of [HA] equilibrium

Smaller version of Ka pKa = -log(Ka) So, Ka = 10^-pKa

Kw = [H+] [OH-] Kw is 10^-14. So we do Kw divided by the concentration of OH- and then plug into -log

Equivalence point = pH7 Phenolphthalein or methyl orange

Equivalence point = roughly pH 5 (below 7) Methyl orange

Equivalence point = roughly 9 (above 7) Phenolphthalein This is because the colour change will happen within the given pH’s of the vertical section

Can work out Ka by using the half equivalence point present within the buffer zone. Half equivalence point is half of the volume at the normal equivalence point At half equivalence point pH = pKa So concentration of weak acid equals concentration of salt: [A-] = [HA]. According to Henderson Hasselback (pH = pKa + log([A-]/[HA]), if A- and HA are equal, then pH = pKa

No clear equivalence point so no suitable indicator for weak acid- use pH metere instead Indicator will change colour very gradually as a weak acid is reacting with a strong base so there will be no clear end point colour change

Acidic buffer: Mixing a weak acid with its conjugate base is- eg: Ethanoic acid and ethanoate ion Alkali buffer: Mixing a weak base with its conjugate acid - ammonia and ammonium chloride Mixing excess weak acid with a strong base

A solution that resists changes to pH when a small amount of acid or base is added HA ⇌ H+ + A− If acid (H+) is added: A− removes it by forming more HA. If base (OH−) is added: HA donates H+ to neutralise OH−.

HA 🔄 H+ + A- A- combined with added H+ to form HA Equilibrium shifts left, reducing H+. HA increases and A- decreases As there are large reserves of A- and HA, so RATIO of concentration remains relatively the same so pH change is minimal

HA 🔄 H+ + A- H+ combines with added OH- to form H20 HA disassociates more to replace lost H+ Equilibrium shifts right, replacing H+ ions that reacted with OH- so HA decreases and A- increases As there are large reserves of A- and HA, the concentrations don’t change much so RATIO remains relatively the same so pH change is minimal

Blood pH is controlled by a hydrogencarbonate buffer system: H2CO3 ⇌ H+ + HCO3− If extra CO2 in blood, CO2 dissolves in blood to form carbonic acid, so concentration of H2CO3 increases This means equilibrium shifts right to counteract change and produce more H+ and HCO3- Large reservoir of excess HCO3- combines with H+ to help control blood pH This maintains blood pH at around 7.4

pH = pKa + log([A-]/[HA]) –> used for working out the pH of buffer calculations involving weak acids and their conjugate base

Dissociation of the weak acid is small enough that the amount of acid lost is negligible.This allows us to make a key approximation: the equilibrium concentration of the acid is virtually the same as its initial concentration. If the dissascoaiton was significant, concentration of HA would decrease Initial [H⁺] from water is negligible → We assume all H⁺ comes from the acid

CH₂ClCOOH is stronger than CH₃COOH because the chlorine atom withdraws electron density from the OH bond which weakens the O-H bond, therefore stabilising the conjugate base. More stable conjugate base (of CH2ClCOO-). So CH₂ClCOOH is a stronger acid

Lower pKa indicates a stronger acid Stronger acid → more dissociation → higher Kₐ → lower pKₐ

a conjugate base