pH, Ka, Kb and Percent Ionization
A swimming pool, a glass of lemon juice, and the drain cleaner under your kitchen sink have one thing in common: each can be described by a single number, the pH. Behind that number is an ion concentration that changes by many orders of magnitude, and pH is simply a convenient way to handle it. By the end of this page you will be able to explain why pool water sits near pH 7.4, lemon juice near 2, and drain cleaner near 14.
In this unit we treat concentrations in mol/L as a good stand-in for activities, which works for fairly dilute solutions. Unless stated otherwise, the temperature is 25 °C, where $K_w = 1.0\times10^{-14}$. If a strong acid or base is so dilute that it contributes about $10^{-7}\ \text{M}$ or less, you cannot ignore the autoionization of water.
pH and pOH: Measuring Acidity
A sensor in the lab gives these values: cleaning-grade hydrochloric acid has $[\text{H}_3\text{O}^+] \approx 1\ \text{M}$, black coffee about $10^{-5}\ \text{M}$, distilled water $10^{-7}\ \text{M}$, and household bleach about $10^{-13}\ \text{M}$. Exponents that different are awkward to compare. How can we squeeze thirteen orders of magnitude into a small, friendly scale?
In the chemistry lab 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…
In the chemistry lab 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 behavior 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: Complete Dissociation
Connect a light bulb to two electrodes dipped in 0.10 M HCl and it glows brightly. Put the same electrodes in 0.10 M $\text{CH}_3\text{COOH}$ (the acid in vinegar) and you get only a faint glow. Both solutions have the same acid concentration, so why does one conduct so much better?
Weak acids and bases only dissociate partially. They reach an equilibrium between the molecule and its ions, described by a constant.
Ammonia, $\mathrm{NH_3}$, acts in aqueous solution as a weak base with a base dissociation constant $K_\mathrm{b} = 1.8\times 10^{-5}$ at $25\,^\circ\mathrm{C}$.…
Ammonia, $\mathrm{NH_3}$, acts in aqueous solution as a weak base with a base dissociation 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 ion-product constant 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 autoionization 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 ionization 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 behavior. Use the ideal model and neglect the autoprotolysis of water where that approximation is valid.
Percent Ionization and Approximations
How much of a weak acid actually ionizes? We measure that with the percent ionization.
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 ionization 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.
Two aqueous solutions at $25\,^\circ\mathrm{C}$, labeled A and B, are available in the lab, both of concentration $0.050\,\mathrm{mol\times L^{-1}}$. Solution A contains…
Two aqueous solutions at $25\,^\circ\mathrm{C}$, labeled A and B, are available in the lab, 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 Working Backward
Add water to a vinegar solution and a conductivity probe shows something curious: the fraction of acid molecules that are ionized goes up. Why would diluting a weak electrolyte push more of it into ions?
Working backward from a measured pH
If you measure the pH of a solution of known concentration, you can solve for $K_a$ instead of using it as given. Convert pH to $[\text{H}_3\text{O}^+] = x$, build the ICE table, and compute $K_a = x^2/(c_0 - x)$. Then $\text{p}K_a = -\log K_a$. If the concentration comes from a mass, first find moles with the molar mass.
If a titration gives the moles of acid (for example, moles of NaOH at the equivalence point for a monoprotic acid), you can also find the molar mass: molar mass = grams of acid / moles of acid.
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 ionization using Le Chatelier's principle or Ostwald's law. Use the ideal model and neglect the autoprotolysis of water where that approximation is valid.
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 neutralization 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.
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 neutralization 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.
Back to the start: Lemon juice has a pH near 2 ($[\text{H}_3\text{O}^+] \approx 10^{-2}\ \text{M}$), pool water is kept near pH 7.4 so it doesn't irritate your eyes, and drain cleaner sits near pH 14 ($[\text{OH}^-] \approx 1\ \text{M}$). From lemon juice (pH 2) to drain cleaner (pH 14) that is twelve pH units, or twelve orders of magnitude in $[\text{H}_3\text{O}^+]$, neatly summarized by a scale that runs from 0 to 14.