Topic 14: Acid-base Equilibria

Here HA donates H$^+$ and is the acid. Water accepts H$^+$ and is the base. Writing only $\ce{HA <=> H+ + A-}$ hides the accepting species, so use water when the question asks how the acid behaves in aqueous solution.

A species can play different roles in different reactions. For example, $\ce{HCO3-}$ can accept a proton to form $\ce{H2CO3}$ or donate one to form $\ce{CO3^2-}$; classify the role from the actual proton transfer.

Match formulae first, then check charge: loss of H$^+$ makes the charge one unit more negative; gain makes it one unit more positive. The two members of a pair normally appear on opposite sides of the equation.

The concentration $[\ce{H+}]$ (or $[\ce{H3O+}]$) is in mol dm$^{-3}$. Because the scale is logarithmic, a decrease of one pH unit corresponds to a tenfold increase in hydrogen-ion concentration.

Use the equilibrium hydrogen-ion concentration in mol dm$^{-3}$, enter its base-10 logarithm, then change the sign. For $[\ce{H+}]=2.50\times10^{-3}$ mol dm$^{-3}$, $\mathrm{pH}=-\log_{10}(2.50\times10^{-3})=2.602$.

Keep unrounded concentration values during a multi-stage calculation. A pH commonly has as many decimal places as the concentration has significant figures when the data justify that precision.

For pH 1.125, $[\ce{H+}]=10^{-1.125}=7.50\times10^{-2}$ mol dm$^{-3}$. The operation is the inverse of the base-10 logarithm.

When pH data are used in dilution, convert both pH values to concentrations before applying conservation of moles or $c_1V_1=c_2V_2$. A pH increase of 1 means a tenfold decrease in $[\ce{H+}]$, not a decrease of 1 mol dm$^{-3}$.

The first dissociation of sulfuric acid is effectively complete, but its second dissociation is an equilibrium. Therefore a 0.100 mol dm$^{-3}$ solution need not contain exactly 0.200 mol dm$^{-3}$ H$^+$.

For 0.500 mol dm$^{-3}$ HCl, $[\ce{H+}]=0.500$ mol dm$^{-3}$ and pH $=0.301$. For 1.25 mol dm$^{-3}$ HCl, pH $=-0.097$: negative pH is mathematically possible.

If acid and base are mixed, calculate reacting moles first, identify the excess H$^+$, divide by the total volume, then calculate pH.

Place equilibrium concentrations of products over reactant and use square brackets. Liquid water is omitted because its effective concentration is constant. For ethanoic acid, replace HA and A$^-$ with $\ce{CH3COOH}$ and $\ce{CH3COO-}$, preserving formulae and charges.

At a fixed temperature, a larger $K_a$ means the dissociation equilibrium lies further toward ions and the weak acid is stronger.

For a weak monoprotic acid of initial concentration $c$, assume dissociation is small so $c-x\approx c$. Then $x\approx\sqrt{K_ac}$ and $\mathrm{pH}=-\log_{10}x$. Convert $pK_a$ first with $K_a=10^{-pK_a}$.

For $c=0.100$ mol dm$^{-3}$ and $pK_a=4.88$, $K_a=1.32\times10^{-5}$ mol dm$^{-3}$, $[\ce{H+}]=1.15\times10^{-3}$ mol dm$^{-3}$ and pH $=2.94$.

In neutral water, $[\ce{H+}]=[\ce{OH-}]=\sqrt{K_w}$. At 25 °C, $K_w=1.00\times10^{-14}$ mol$^2$ dm$^{-6}$, so neutral pH is 7.00.

Kw changes with temperature. If $K_w=5.5\times10^{-14}$ mol$^2$ dm$^{-6}$ at 50 °C, neutral water has pH about 6.6 because the two ion concentrations remain equal.

At 25 °C, 0.200 mol dm$^{-3}$ $\ce{Ba(OH)2}$ gives $[\ce{OH-}]=0.400$ mol dm$^{-3}$. Thus pOH $=0.398$ and pH $=14.000-0.398=13.602$.

For acid-base mixtures, use balanced reacting moles, calculate excess OH$^-$ concentration in the total volume, and only then convert to pH.

The inverse conversions are $K_a=10^{-pK_a}$ and $K_w=10^{-pK_w}$. Because of the negative logarithm, a smaller $pK_a$ corresponds to a larger $K_a$ and hence a stronger weak acid.

If $K_a=1.38\times10^{-4}$ mol dm$^{-3}$, $pK_a=3.86$. At 25 °C, $K_w=1.00\times10^{-14}$ mol$^2$ dm$^{-6}$ gives $pK_w=14.00$.

A tenfold dilution of a strong monoprotic acid makes $[\ce{H+}]$ ten times smaller, so pH rises by 1. A weak acid dissociates to a greater fraction after dilution, partly replacing the removed H$^+$; its pH therefore rises by less than 1 in the supplied comparison.

For the same concentration series, HCl pH may rise 1.00 → 2.00 → 3.00 on successive tenfold dilutions, while ethanoic acid might rise 2.88 → 3.38 → 3.88.

If a solution has $c=0.500$ mol dm$^{-3}$ and pH 1.20, then $x=0.0631$ mol dm$^{-3}$ and $K_a=(0.0631)^2/(0.500-0.0631)=9.11\times10^{-3}$ mol dm$^{-3}$.

State the relevant chemistry: one H$^+$ and one conjugate-base ion form per dissociated acid molecule, and any later dissociation is negligible when the question says so.

The equivalence volume is where stoichiometric acid and base amounts have reacted. For a diprotic acid titrated by a strong base, the second equivalence volume is twice the first when both protons react sequentially under the same conditions.

Use the supplied Data Booklet range or approximately $pK_{In}\pm1$, and state the endpoint colour if requested. A named indicator earns justification only when its range is compared with the actual graph.

A typical acidic buffer contains appreciable amounts of a weak acid and its conjugate base; an alkaline buffer contains a weak base and its conjugate acid. Both components are needed to consume the two kinds of added reagent.

Because HA and A$^-$ are both present as a large reservoir, removing a small added amount changes their concentration ratio only slightly. The corresponding $[\ce{H+}]$ and pH therefore change only slightly.

The same logic applies to $\ce{NH3/NH4+}$: ammonia accepts added H$^+$ to form $\ce{NH4+}$, while $\ce{NH4+}$ supplies acid capacity against added OH$^-$.

First identify the conjugate pair. If strong acid or base has partly neutralised a weak component, calculate reacting moles and the remaining HA/A$^-$ amounts before using the equilibrium expression. When both share the same final volume, their mole ratio may replace the concentration ratio.

A buffer contains 0.175 mol HA and 0.100 mol A$^-$ with $K_a=1.70\times10^{-5}$. Then $[\ce{H+}]=1.70\times10^{-5}(0.175/0.100)=2.98\times10^{-5}$ mol dm$^{-3}$ and pH $=4.53$.

Convert the target pH and supplied $K_a$ or $pK_a$ into the required conjugate-base:weak-acid ratio. Use the known acid concentration or moles to find the required salt concentration, moles, mass or volume ratio.

For $pK_a=3.86$ and target pH 3.71, $[\ce{A-}]/[\ce{HA}]=10^{-0.15}=0.708$. If $[\ce{HA}]=1.55$ mol dm$^{-3}$ in 1.00 dm$^3$, 1.10 mol of conjugate base is required when volume change is neglected.

The ratio sets pH, while the total amounts help determine buffer capacity. Follow any stated assumption about unchanged volume and use molar mass if a solid salt mass is requested.

Read the equivalence volume from the centre of the steep section. Halve that volume, read the pH at this half-neutralisation point, then take $pK_a=$ that pH and $K_a=10^{-pK_a}$.

For a resolved diprotic curve, each dissociation has its own half-equivalence point midway between its neighbouring equivalence volumes, giving a separate $pK_a$.

Cells and blood require pH within a limited range for biochemical processes. Added H$^+$ is consumed by $\ce{HCO3-}$ to form $\ce{H2CO3}$; carbonic acid can form $\ce{CO2}$ and water, linking the buffer to carbon dioxide removal.

A weak acid and its salt can buffer food against pH changes caused by bacterial or fungal activity. For example, citric acid with citrate helps limit pH drift in a food such as marmalade, slowing deterioration associated with that change.

Repeat pH readings or the titration, use smaller additions near the steep section, and keep temperature constant because equilibrium constants and electrode response depend on temperature.

If a known-concentration weak-acid solution is measured directly, calculate $[\ce{H+}]$ from pH and use $K_a=x^2/(c-x)$. State which experimental route and assumptions are being used.