pH, Ka, Kb and degree of dissociation
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.
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.
Strong acids and bases: full dissociation
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.
Ammonia, $\mathrm{NH_3}$, acts in aqueous solution as a weak base with a basicity constant $K_\mathrm{b} = 1.8\times 10^{-5}$ at $25\,^\circ\mathrm{C}$. Consider the…
Ammonia, $\mathrm{NH_3}$, acts in aqueous solution as a weak base with a basicity constant $K_\mathrm{b} = 1.8\times 10^{-5}$ at $25\,^\circ\mathrm{C}$. Consider the proton transfer equilibrium with water for the species ammonia and for its conjugate acid, the ammonium ion ($\mathrm{NH_4^+}$). The ionic product of water is $K_\mathrm{w} = 1.0\times 10^{-14}$ at that temperature. a) Write the acid dissociation reaction of the ammonium ion in water and deduce the mathematical relationship between $K_\mathrm{a}(\mathrm{NH_4^+})$, $K_\mathrm{b}(\mathrm{NH_3})$ and $K_\mathrm{w}$. b) Calculate the numerical value of $K_\mathrm{a}$ for the ammonium ion and justify which of the two species reacts to a greater extent with water at equal initial concentration, by comparing their respective constants. Use the ideal model and neglect the autoprotolysis of water where that approximation is valid.
You have an aqueous solution of methylamine, $\mathrm{CH_3NH_2}$, of concentration $0.050\,\mathrm{mol\times L^{-1}}$ at $25\,^\circ\mathrm{C}$. The basicity constant of…
You have an aqueous solution of methylamine, $\mathrm{CH_3NH_2}$, of concentration $0.050\,\mathrm{mol\times L^{-1}}$ at $25\,^\circ\mathrm{C}$. The basicity constant of methylamine is $K_\mathrm{b} = 4.4\times 10^{-4}$ at that temperature and the ionic product of water is $K_\mathrm{w} = 1.0\times 10^{-14}$. a) Set out the ionisation equilibrium in water, express the constant $K_\mathrm{b}$ in terms of the initial concentration $c_0$ and the degree of dissociation $\alpha$, and check quantitatively whether the approximation $1 - \alpha \approx 1$ is applicable using the $5\%$ criterion. b) Calculate the value of the degree of dissociation $\alpha$, the concentration of all species at equilibrium and the $\mathrm{pH}$ of the solution, assuming ideal behaviour. Use the ideal model and neglect the autoprotolysis of water where that approximation is valid.
Degree of dissociation and approximations
How much of a weak acid actually breaks up? We answer with the degree of dissociation.
In the laboratory, a volumetric flask is prepared with $100\,\mathrm{mL}$ of an aqueous solution of a weak monoprotic acid $\mathrm{HA}$ of concentration…
In the laboratory, a volumetric flask is prepared with $100\,\mathrm{mL}$ of an aqueous solution of a weak monoprotic acid $\mathrm{HA}$ of concentration $0.10\,\mathrm{mol\times L^{-1}}$, whose acidity constant is $K_\mathrm{a} = 1.8\times 10^{-5}$ at $25\,^\circ\mathrm{C}$. This sample is then transferred in its entirety to another flask and diluted with distilled water to a final volume of $1000\,\mathrm{mL}$. The diagram of the dilution process and the associated parameters are shown below:
. At $25\,^\circ\mathrm{C}$ and assuming an ideal solution with $\mathrm{pH} = -\log[\mathrm{H_3O^+}]$, determine: a) The degree of dissociation, $\alpha_1$, and the $\mathrm{pH}$ of the initial concentrated solution. b) The degree of dissociation, $\alpha_2$, and the $\mathrm{pH}$ of the resulting dilute solution, justifying the effect of dilution on the degree of ionisation using Le Chatelier's principle or Ostwald's law. Use the ideal model and neglect the autoprotolysis of water where that approximation is valid.
Chloroacetic acid, $\mathrm{CH_2ClCOOH}$, is used in organic synthesis. At $25\,^\circ\mathrm{C}$, an aqueous solution of this acid with a concentration…
Chloroacetic acid, $\mathrm{CH_2ClCOOH}$, is used in organic synthesis. At $25\,^\circ\mathrm{C}$, an aqueous solution of this acid with a concentration $c_0 = 0.010\,\mathrm{mol\times L^{-1}}$ has an acidity constant $K_\mathrm{a} = 1.4\times 10^{-3}$. a) Write the chemical equation for the ionisation equilibrium and calculate the concentration of $[\mathrm{H_3O^+}]$ by solving the full quadratic equation. State explicitly the percentage error in $[\mathrm{H_3O^+}]$, relative to the value from the quadratic solution, that would be made if the dissociated amount were neglected compared with the initial concentration. b) Determine the $\mathrm{pH}$ of the solution and the percentage of acid that remains in molecular, undissociated form at equilibrium. Use the ideal model and neglect the autoprotolysis of water where that approximation is valid.
A pure sample of $1,220\,\mathrm{g}$ of an unknown solid weak monoprotic acid, $\mathrm{HA}$, is dissolved in distilled water to a total volume of $250\,\mathrm{mL}$ at…
A pure sample of $1,220\,\mathrm{g}$ of an unknown solid weak monoprotic acid, $\mathrm{HA}$, is dissolved in distilled water to a total volume of $250\,\mathrm{mL}$ at $25\,^\circ\mathrm{C}$. The $\mathrm{pH}$ of the resulting solution is measured experimentally with a calibrated pH meter, giving a value of $3.11$. In a separate total neutralisation experiment, it is found that the whole $250\,\mathrm{mL}$ of this solution requires exactly $20.0\,\mathrm{mL}$ of standard solution of $\mathrm{NaOH}$ $0.500\,\mathrm{mol\times L^{-1}}$ to reach the stoichiometric equivalence point. Assuming an ideal solution with $\mathrm{pH} = -\log[\mathrm{H_3O^+}]$, determine: a) The molar mass of the unknown weak monoprotic acid. b) The acid dissociation constant $K_\mathrm{a}$ of the acid and its degree of dissociation $\alpha$ in the prepared solution, and justify whether the classical equilibrium approximation was valid. Use the ideal model and neglect the autoprotolysis of water where that approximation is valid.
Two aqueous solutions at $25\,^\circ\mathrm{C}$, labelled A and B, are available in the laboratory, both of concentration $0.050\,\mathrm{mol\times L^{-1}}$. Solution A…
Two aqueous solutions at $25\,^\circ\mathrm{C}$, labelled A and B, are available in the laboratory, both of concentration $0.050\,\mathrm{mol\times L^{-1}}$. Solution A contains formic acid, $\mathrm{HCOOH}$ ($K_\mathrm{a} = 1.8\times 10^{-4}$), and solution B contains acetic acid, $\mathrm{CH_3COOH}$ ($K_\mathrm{a} = 1.8\times 10^{-5}$). Consider ideal solutions with $K_\mathrm{w} = 1.0\times 10^{-14}$. a) Calculate the percentage degree of dissociation and the $\mathrm{pH}$ of both solutions, justifying which will conduct electric current better. b) If $0.0050\,\mathrm{mol}$ of solid sodium acetate ($\mathrm{CH_3COONa}$, a strong electrolyte) is added, with no change in volume, to $100\,\mathrm{mL}$ of solution B, calculate the new value of $\mathrm{pH}$ and explain the change in the degree of dissociation using Le Chatelier's principle. To compare conductivity, consider the contribution of the hydronium ion to be predominant and the mobilities of the two anions to be comparable. Use the ideal model and neglect the autoprotolysis of water where that approximation is valid.
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.
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.