The water in a municipal swimming pool, the lemon juice on your pancakes and the drain cleaner under the sink all have one number in common: pH. It sums up how acidic or basic a solution is. Behind that number lies a concentration of ions that can vary over fourteen powers of ten, and pH is simply a handy way to manage it.
In this section we use concentrations in mol L⁻¹ in place of activities. This is fine for dilute solutions. We work at 25 °C unless told otherwise, and we only consider dilute aqueous solutions. Mixtures of acids and bases are not required for the syllabus.
pH: measuring acidity
Measured with a pH sensor, a strong laboratory acid (pH 0) has $[\mathrm{H^+}] \approx 1\ \text{mol L}^{-1}$, black coffee about $10^{-5}\ \text{mol L}^{-1}$, pure water $10^{-7}\ \text{mol L}^{-1}$ and a strongly basic drain cleaner (pH 14) only about $10^{-14}\ \text{mol L}^{-1}$. Writing such different negative powers of ten is awkward. How can we squeeze fourteen orders of magnitude into one easy scale?
For a base you are usually given $[\mathrm{OH^-}]$ rather than $[\mathrm{H_3O^+}]$. Use $K_\mathrm{w}$ to convert: $[\mathrm{H_3O^+}] = K_\mathrm{w}/[\mathrm{OH^-}]$, then take the pH as usual. This is the same relationship as before, just rearranged.
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In the chemistry laboratory there is an aqueous solution of nitric acid, $\mathrm{HNO_3}$, of concentration $0.025\,\mathrm{mol\times L^{-1}}$ at $25\,^\circ\mathrm{C}$.…
In the chemistry laboratory there is an aqueous solution of nitric acid, $\mathrm{HNO_3}$, of concentration $0.025\,\mathrm{mol\times L^{-1}}$ at $25\,^\circ\mathrm{C}$. Given that nitric acid behaves as a strong monoprotic acid that dissociates completely in water and that the ionic product of water is $K_\mathrm{w} = 1.0\times 10^{-14}$ at that temperature, assume ideal behaviour where $\mathrm{pH} = -\log[\mathrm{H_3O^+}]$. Calculate: a) The concentration of hydronium ions, $[\mathrm{H_3O^+}]$, and the value of the $\mathrm{pH}$ of the solution. b) The concentration of hydroxide ions, $[\mathrm{OH^-}]$, present in the medium. Use the ideal model and neglect the autoprotolysis of water where that approximation is valid.
An aqueous solution is prepared by dissolving $0.171\,\mathrm{g}$ of barium hydroxide, $\mathrm{Ba(OH)2}$, in distilled water to make an exact volume of…
An aqueous solution is prepared by dissolving $0.171\,\mathrm{g}$ of barium hydroxide, $\mathrm{Ba(OH)_2}$, in distilled water to make an exact volume of $500\,\mathrm{mL}$ at $25\,^\circ\mathrm{C}$. This compound is assumed to be a strong base that dissociates completely according to the process $\mathrm{Ba(OH)_2(s) \rightarrow Ba^{2+}(aq) + 2\,OH^-(aq)}$. Molar mass of $\mathrm{Ba(OH)_2}$: $171.3\,\mathrm{g\times mol^{-1}}$; $K_\mathrm{w} = 1.0\times 10^{-14}$. Under the ideal approximation, determine: a) The molar concentration of $\mathrm{OH^-}$ ions in the resulting solution. b) The $\mathrm{pOH}$ and the $\mathrm{pH}$ of the final solution. Use the ideal model and neglect the autoprotolysis of water where that approximation is valid.
Connect a bulb to two electrodes in 0.10 mol L⁻¹ HCl and it glows brightly. In 0.10 mol L⁻¹ ethanoic acid it barely glimmers. Both solutions contain the same amount of acid, so why does one conduct so much better?
Weak acids and bases dissociate only partly. An equilibrium is set up between the molecules and their ions. The constants $K_\mathrm{a}$ and $K_\mathrm{b}$ below are usually studied in Sixth Year of the Leaving Certificate course.
Degree of dissociation and approximations
How much of a weak acid actually breaks up? We answer with the degree of dissociation.
Dilution, mixtures and inverse problems
When you dilute ethanoic acid, the fraction of its molecules that are ionised increases (the degree of dissociation, α, rises), even though there are fewer ions per litre. Why does diluting a weak electrolyte make its molecules split into ions more easily?
Mixtures of a strong acid and a strong base are an extension beyond the syllabus, so the exercise on mixtures below is extension work. Work in moles: subtract the moles of $\mathrm{H^+}$ and $\mathrm{OH^-}$, divide what is left by the total volume and then find the pH.
Practice★★★Not started
A laboratory reactor is filled with a mixture of $250\,\mathrm{mL}$ of an aqueous solution of hydrochloric acid, $\mathrm{HCl}$, of concentration…
A laboratory reactor is filled with a mixture of $250\,\mathrm{mL}$ of an aqueous solution of hydrochloric acid, $\mathrm{HCl}$, of concentration $0.060\,\mathrm{mol\times L^{-1}}$ and $150\,\mathrm{mL}$ of an aqueous solution of barium hydroxide, $\mathrm{Ba(OH)_2}$, of concentration $0.040\,\mathrm{mol\times L^{-1}}$ at $25\,^\circ\mathrm{C}$. Assume that the volumes are additive (final volume $400\,\mathrm{mL}$), that both substances behave as strong electrolytes that are fully dissociated, and that $K_\mathrm{w} = 1.0\times 10^{-14}$ at that temperature in an ideal solution. a) Write the net ionic neutralisation reaction and determine, with justification, which reagent is in excess, quantifying the moles remaining. b) Calculate the molar concentration of all the ionic species present in the resulting mixture ($\mathrm{H_3O^+}$, $\mathrm{OH^-}$, $\mathrm{Ba^{2+}}$ and $\mathrm{Cl^-}$) and determine the $\mathrm{pH}$ of the final solution. Use the ideal model and neglect the autoprotolysis of water where that approximation is valid.
Back to the start: Lemon juice has a pH of about 2 ($[\mathrm{H_3O^+}] \approx 10^{-2}\ \text{mol L}^{-1}$). Pool water is kept at about pH 7.4 so it does not irritate your eyes. Drain cleaner is near pH 14 ($[\mathrm{OH^-}] \approx 1\ \text{mol L}^{-1}$). Between lemon juice and drain cleaner that is twelve powers of ten, and the whole range of fourteen (from $1$ to $10^{-14}\ \text{mol L}^{-1}$) is neatly squeezed into a scale from 0 to 14.
Summary
Let's practice! 0 medium exercises in this section