Grade 11

Electrolytes, pH, buffer solutions

How dissolved ions conduct electricity, how acid–base equilibria set pH, and how conjugate pairs resist pH change in a buffer.

1 Lower secondary 2 Upper secondary 3 Undergraduate 4 Advanced 5 Research

IntuitionWater can carry current when it contains mobile ions

Some dissolved substances remain neutral molecules; others form ions. Strong electrolytes produce ions almost completely, whereas weak electrolytes establish a balance between ions and intact molecules. The balance also explains why a weak acid can set a pH and why a buffer resists change.

In the particle view, most grey HA molecules remain undissociated, while a smaller, paired population of H₃O⁺ and A⁻ is present. The temperature control changes the simulated thermal motion; particle counts are illustrative, not a quantitative equilibrium measurement.

Grey HA molecules remain mostly intact; a few red H₃O⁺ and blue A⁻ ions show partial dissociation. This is illustrative, not to scale; the temperature slider changes simulated motion.

At half-equivalence, equal amounts of HA and A⁻ give a buffer plateau with pH = pKa.

SchoolElectrolytes, acid–base equilibria, and pH

Definition: Electrolyte and dissociation

An electrolyte forms mobile ions in solution or on melting and therefore conducts electricity; a non-electrolyte does not form appreciable ions. In Arrhenius language, acids increase aqueous H₃O⁺ and bases increase OH⁻. Strong acids/bases ionise essentially completely in dilute water; weak ones ionise only partly.

Typical strong acids include HCl, HBr, HI, HNO₃, HClO₄ and the first dissociation of H₂SO₄; common strong bases are soluble Group 1 hydroxides and Ca(OH)₂, Sr(OH)₂ and Ba(OH)₂. Acetic acid, HF, HCN and NH₃ are weak. Conductivity depends on both ion concentration and mobility, not simply on whether a substance is labelled an acid.

\ce{HA + H2O <=> H3O+ + A-}\\K_a=\frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]},\qquad \alpha=\frac{x}{C}

The Brønsted–Lowry definition is broader: an acid donates a proton and a base accepts one. Each acid–base reaction has conjugate pairs differing by one proton, such as HA/A⁻ and H₃O⁺/H₂O. At 25 °C, water autoionisation has $K_w=[H⁺][OH⁻]=1.0×10⁻¹⁴$ ; thus pH + pOH = 14.00.

\mathrm{pH}=-\log_{10}a_{\ce{H+}}\approx-\log_{10}[\ce{H+}],\qquad \mathrm{pOH}=-\log_{10}[\ce{OH-}]\\K_aK_b=K_w\quad\text{for a conjugate pair at the same temperature}

Selected acid dissociation constants at 25 °C
Species	Strength or pKa	Value
HCl, HNO₃	Strong acids	Essentially complete
CH₃COOH	Weak acid	Ka = 1.8×10⁻⁵; pKa = 4.74
HF	Weak acid	Ka = 6.8×10⁻⁴
HCN	Weak acid	Ka = 6.2×10⁻¹⁰
NH₄⁺	Conjugate acid of NH₃	Ka = 5.6×10⁻¹⁰
H₂CO₃	Diprotic acid	pKa₁ = 6.35; pKa₂ = 10.33

Example: Strong-acid pH

Find the pH of 0.010 M HCl at 25 °C.

Solution

HCl is effectively fully dissociated, so [H⁺] ≈ 0.010 M and pH = −log(0.010) = 2.00.

Example: Weak-acid pH: exact and approximate

Find the pH of 0.10 M acetic acid, Ka = 1.8×10⁻⁵.

Solution

Let x=[H⁺]. The small-dissociation estimate is x≈√(KaC)=1.34×10⁻³ M, α≈1.34%, so α<5% and the approximation is justified; pH≈2.87. Exact mass-action gives Ka=x²/(C−x), hence x=[−Ka+√(Ka²+4KaC)]/2=1.33×10⁻³ M and pH=2.875≈2.87.

Ostwald’s dilution law follows from the mass-action expression: for a monoprotic weak acid, Ka=Cα²/(1−α). Dilution generally increases α, but lowers the concentration of ions. A common ion such as A⁻ shifts HA dissociation toward HA and suppresses ionisation.

Approximate pH of familiar aqueous materials (variable by sample)
Material	Approximate pH
Stomach contents	1.5–2
Coffee	≈5
Blood	≈7.4
Seawater	≈8.1
Household bleach	≈12.5

UndergraduateQuantitative acid–base systems and buffers

A buffer contains a weak acid HA and its conjugate base A⁻ (often supplied as a soluble salt). Added H⁺ is consumed by A⁻ to form HA; added OH⁻ is consumed by HA to form A⁻ and water. The solution therefore changes pH less than an unbuffered solution, until one component is depleted.

\mathrm{pH}=\mathrm{p}K_a+\log_{10}\frac{[\ce{A-}]}{[\ce{HA}]}\\\text{use concentrations (or mole amounts in the same final volume) for an ideal dilute buffer}

Example: Acetate buffer and half-equivalence

Find pH for 0.10 M CH₃COOH/0.10 M CH₃COONa, then after adding 0.010 mol NaOH to 1.0 L of this buffer. Use pKa=4.74.

Solution

Initially [A⁻]=[HA], so pH=pKa=4.74 (the half-equivalence condition). The strong base converts 0.010 mol HA into A⁻: amounts become 0.11 and 0.09 mol in the same final volume. Henderson–Hasselbalch gives 4.74+log(0.11/0.09)=4.827≈4.83.

Example: Preparing a target acetate buffer

How much 0.10 M NaOH is needed to convert 50.0 mL of 0.10 M acetic acid into a buffer of pH 4.74? Assume additive volumes are negligible.

Solution

At pH=pKa the ratio A⁻/HA is 1, so neutralise half of the initial 5.00 mmol acid: 2.50 mmol OH⁻. Volume=2.50 mmol/(0.10 mmol mL⁻¹)=25.0 mL. The remaining 2.50 mmol HA and formed 2.50 mmol A⁻ make the buffer.

The buffer capacity is greatest near pH=pKa, where both members are appreciable, and the useful range is approximately pKa±1. Blood uses the H₂CO₃/HCO₃⁻ system to maintain pH near 7.4; its apparent pKa in physiological conditions is about 6.1. Acetate buffers are convenient laboratory examples. Diluting a buffer leaves the ratio—and approximately its pH—similar but lowers its capacity.

AdvancedActivity, charge balance, and polyprotic systems

Thermodynamic pH is −log a(H⁺), with activity aᵢ=γᵢ[i]/c°. Concentration-based classroom expressions assume γ≈1. In the Debye–Hückel limiting law at 25 °C, log₁₀γᵢ≈−0.51zᵢ²√I (I in mol L⁻¹), where I=½Σcᵢzᵢ². This limiting expression is reliable only at low ionic strength.

K_a=\frac{a_{\ce{H+}}a_{\ce{A-}}}{a_{\ce{HA}}},\qquad I=\frac12\sum_i c_i z_i^2,\qquad \log_{10}\gamma_i\approx-0.51z_i^2\sqrt I

For a polyprotic acid HₙA, solve mass balances, each Ka relation, water equilibrium and electroneutrality together. For a diprotic acid with total analytical concentration C_T, [H₂A]+[HA⁻]+[A²⁻]=C_T and [H⁺]=[OH⁻]+[HA⁻]+2[A²⁻] (add spectator-ion charges when salts are present). The fractional distributions are α₀=[H⁺]²/D, α₁=Ka₁[H⁺]/D, α₂=Ka₁Ka₂/D, D=[H⁺]²+Ka₁[H⁺]+Ka₁Ka₂; these generate α-diagrams versus pH.

For an ideal monoprotic buffer of total analytical concentration C=[HA]+[A⁻], define α=[A⁻]/C. Its buffer capacity per unit volume is β=2.303 Cα(1−α), maximised at α=½, equivalently pH=pKa. Water, added strong acid/base and non-ideal activities add terms to measured capacity.

Søren Sørensen introduced the pH notation in 1909 while studying acidity in biochemical systems; the logarithmic scale made widely differing hydrogen-ion levels easier to compare.

References

Inorganic Chemistry · Catherine E. Housecroft; Alan G. Sharpe, 2018
Chemistry: The Central Science · Theodore L. Brown et al., 2018