--- title: "Acids and Bases: Ka and pKa" book: "School Chemistry — Grades 1 to 12" subject: chemistry language: en chapter: 44 exercises: 15 source: https://one-course.com/books/chemistry/1/en/chapter/44-acids-and-bases-ka-and-pka license: CC-BY-NC-SA-4.0 credit: "One Chemistry Book, One Course (one-course.com)" --- # Chapter 44 — Acids and Bases: Ka and pKa Many ants defend themselves with methanoic acid, and on the skin it burns. Methanoic acid is a small [carboxylic acid](https://one-course.com/books/chemistry/1/en/chapter/33-functional-groups-and-families#def-g11-functional-groups-carboxylic-acid). Hydrochloric acid at the same concentration would burn far more. Both are acids; both give a solution of [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) below 7; yet they are not acidic in the same way. To describe an acid fully, two numbers are needed: how much of it there is, its concentration, and how readily it gives away its hydrogen [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion), its [acidity constant](#def-g12-ka-and-pka-ka). **You already know.** Acidic, basic and [neutral solutions](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-acidic); the [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) scale, from 0 to 14, lower when there are more hydrogen [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) ([Chapter 18](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#ch-g9-acids-bases-ph)). The [reaction quotient](https://one-course.com/books/chemistry/1/en/chapter/43-chemical-equilibrium-quotient-and-constant#def-g12-equilibrium-quotient) $Q$ and the [equilibrium constant](https://one-course.com/books/chemistry/1/en/chapter/43-chemical-equilibrium-quotient-and-constant#def-g12-equilibrium-constant) $K$ ([Chapter 43](https://one-course.com/books/chemistry/1/en/chapter/43-chemical-equilibrium-quotient-and-constant#ch-g12-equilibrium)). A [curly arrow](https://one-course.com/books/chemistry/1/en/chapter/40-reaction-mechanisms-curly-arrows#def-g12-curly-arrows-curly-arrow) shows a pair of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) moving ([Chapter 40](https://one-course.com/books/chemistry/1/en/chapter/40-reaction-mechanisms-curly-arrows#ch-g12-curly-arrows)). ![Wood ants: like many ants, they defend themselves with methanoic acid.](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-ka-and-pka/img-0622cdc5eea6.jpg) *Wood ants: like many ants, they defend themselves with methanoic acid.* ## 44.1 Brønsted acids and bases **Definition 44.1 (Brønsted acid and base, acid–base couple).** A *Brønsted acid* is a species able to give a hydrogen [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) $\ce{H+}$; a *Brønsted base* is a species able to take one. An acid HA and the base $\ce{A-}$ it becomes by losing $\ce{H+}$ form an *acid–base couple*, written $\ce{HA}$/$\ce{A-}$: $\ce{HA} \rightleftharpoons \ce{A-} + \ce{H+}$. **Definition 44.2 (Oxonium ion, ampholyte).** In water a hydrogen [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) never stays alone: it is taken by a water [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule), giving the *oxonium ion* $\ce{H3O+}$. A species that is the acid of one couple and the base of another is an *ampholyte*; water is one, acid of the couple $\ce{H2O}$/$\ce{OH-}$ and base of the couple $\ce{H3O+}$/$\ce{H2O}$. **Example 44.3 (An acid in water).** When ethanoic acid [dissolves](https://one-course.com/books/chemistry/1/en/chapter/2-mixing-and-dissolving#def-g2-mixing-and-dissolving-dissolve), a [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) of a water [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) takes the hydrogen of its $\ce{O-H}$, and the $\ce{O-H}$ pair stays on the acid’s oxygen: $$ \ce{CH3COOH + H2O <=> CH3COO- + H3O+} . $$ Two couples exchange a hydrogen [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion): $\ce{CH3COOH}$/$\ce{CH3COO-}$ gives it, $\ce{H3O+}$/$\ce{H2O}$ takes it. From now on, the hydrogen [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) of earlier chapters are written $\ce{H3O+}$ whenever water is the [solvent](https://one-course.com/books/chemistry/1/en/chapter/9-solutions-and-solubility#def-g6-solutions-and-solubility-solute). ![A hydrogen ion passes from the acid to water: a lone pair of water forms the new O-H bond (orange), and the old O-H pair stays on the oxygen of the ethanoate ion (blue).](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-ka-and-pka/fig-016661e8750a.svg) *A hydrogen [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) passes from the acid to water: a [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) of water forms the new $\ce{O-H}$ bond (orange), and the old $\ce{O-H}$ pair stays on the oxygen of the ethanoate [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) (blue).* **History — From Arrhenius to Brønsted, 1887–1923.** In 1887 Svante Arrhenius proposed that acids give hydrogen [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) in water and bases hydroxide [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion). In 1923 Johannes Brønsted, and independently Thomas Lowry, broadened the idea: an acid is any species that gives a hydrogen [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion), a base any species that takes one, in water or not. The ammonia [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule), which contains no hydroxide, became a base like any other. ![Svante Arrhenius (1859–1927).](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-ka-and-pka/img-6d1cbc935e6b.jpg) *Svante Arrhenius (1859–1927).* ## 44.2 The pH, quantitatively **Definition 44.4 (The pH, precisely).** In a dilute [aqueous solution](https://one-course.com/books/chemistry/1/en/chapter/9-solutions-and-solubility#def-g6-solutions-and-solubility-solute), the [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) is defined by $$ \mathrm{pH} = -\log\frac{[\ce{H3O+}]}{c^\circ}, \qquad\text{that is}\qquad [\ce{H3O+}] = c^\circ \times 10^{-\mathrm{pH}} , $$ where $\log$ is the decimal logarithm. **Proposition 44.5 (Tenfold dilution, one unit).** If $[\ce{H3O+}]$ is divided by 10, the [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) increases by 1; if it is multiplied by 10, the [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) decreases by 1. **Proof.** $-\log\big([\ce{H3O+}]/10c^\circ\big) = -\log\big([\ce{H3O+}]/c^\circ\big) + \log 10 = \mathrm{pH} + 1$. ∎ **Definition 44.6 (Strong and weak acids and bases).** A *strong acid* reacts totally with water: none of the acid remains as such in the solution. A *weak acid* reacts partially: its reaction with water reaches an equilibrium. Likewise a *strong base* reacts totally with water, a *weak base* partially. **Proposition 44.7 (pH of a strong acid).** A solution of a [strong acid](#def-g12-ka-and-pka-strong-acid) of concentration $c$ (not too dilute) has $[\ce{H3O+}] = c$ and $\mathrm{pH} = -\log(c/c^\circ)$. **Proof.** The reaction $\ce{HA + H2O -> A- + H3O+}$ is total: each [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) of acid gives one [oxonium ion](#def-g12-ka-and-pka-oxonium). The [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) coming from water itself are negligible as long as $c$ is much larger than $10^{-7}\,\mathrm{mol}/\mathrm{L}$. ∎ **Example 44.8 (Hydrochloric acid).** Hydrochloric acid at $0.010\,\mathrm{mol}/\mathrm{L}$: $[\ce{H3O+}] = 0.010\,\mathrm{mol}/\mathrm{L}$ and $\mathrm{pH} = -\log 0.010 = 2.0$. Diluted ten times, [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 3.0. ## 44.3 Water and the ionic product **Definition 44.9 (Ionic product of water).** Water reacts very slightly with itself, $\ce{2H2O <=> H3O+ + OH-}$. The [equilibrium constant](https://one-course.com/books/chemistry/1/en/chapter/43-chemical-equilibrium-quotient-and-constant#def-g12-equilibrium-constant) of this reaction is the *ionic product of water*: $$ K_e = \frac{[\ce{H3O+}]\,[\ce{OH-}]}{(c^\circ)^2}, \qquad pK_e = -\log K_e . $$ **Proposition 44.10 (Neutral water and strong bases).** At $25\,{}^{\circ}\mathrm{C}$, $K_e = 1.0 \times 10^{-14}$ and $pK_e = 14.0$. In pure water $[\ce{H3O+}] = [\ce{OH-}] = 1.0 \times 10^{-7}\,\mathrm{mol}/\mathrm{L}$: [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 7.0. A [strong base](#def-g12-ka-and-pka-strong-acid) of concentration $c$ gives $[\ce{OH-}] = c$, so $[\ce{H3O+}] = K_e (c^\circ)^2 / c$ and $\mathrm{pH} = 14.0 + \log(c/c^\circ)$. **Proof.** In pure water the two [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) come in pairs, so they are equal, and their product is $1.0 \times 10^{-14}$: each is $1.0 \times 10^{-7}\,\mathrm{mol}/\mathrm{L}$. For the base, $K_e$ still holds in the solution: $-\log([\ce{H3O+}]/c^\circ) = -\log K_e + \log(c/c^\circ)$. ∎ **Example 44.11 (Sodium hydroxide).** Sodium hydroxide at $0.010\,\mathrm{mol}/\mathrm{L}$: $[\ce{OH-}] = 0.010$, so $[\ce{H3O+}] = 1.0 \times 10^{-14}/0.010 = 1.0 \times 10^{-12}\,\mathrm{mol}/\mathrm{L}$ and [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 12.0. ## 44.4 The acidity constant **Definition 44.12 (Acidity constant, pKapK_apKa​).** The *acidity constant* $K_a$ of a couple $\ce{HA}$/$\ce{A-}$ is the [equilibrium constant](https://one-course.com/books/chemistry/1/en/chapter/43-chemical-equilibrium-quotient-and-constant#def-g12-equilibrium-constant) of the reaction of the acid with water, $\ce{HA + H2O <=> A- + H3O+}$: $$ K_a = \frac{[\ce{A-}]\,[\ce{H3O+}]}{[\ce{HA}]\,c^\circ} . $$ The *pKa* of the couple is $pK_a = -\log K_a$. **Proposition 44.13 (The smaller the pKapK_apKa​, the stronger the acid).** Of two [weak acids](#def-g12-ka-and-pka-strong-acid) at the same concentration, the one with the larger $K_a$, that is the smaller $pK_a$, reacts more with water and gives the lower [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph); its conjugate base is the weaker. **Proof.** A larger $K_a$ means more $\ce{A-}$ and $\ce{H3O+}$ at equilibrium for the same $\ce{HA}$. The base $\ce{A-}$ of a couple that gives up its hydrogen [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) easily takes it back reluctantly. ∎ ![Some couples placed by their pK_a at 25\, C, acids on the left, their bases on the right. The lower a couple on the scale, the stronger its acid; the higher, the stronger its base.](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-ka-and-pka/fig-8b2008bb0486.svg) *Some couples placed by their $pK_a$ at $25\,{}^{\circ}\mathrm{C}$, acids on the left, their bases on the right. The lower a couple on the scale, the stronger its acid; the higher, the stronger its base.* **Method 44.14 (pH of a weak acid).** For a [weak acid](#def-g12-ka-and-pka-strong-acid) of concentration $c$ and constant $K_a$: 1. Write the [progress table](https://one-course.com/books/chemistry/1/en/chapter/27-the-reaction-progress-table#def-g10-reaction-progress-table-extent) of $\ce{HA + H2O <=> A- + H3O+}$ per litre, with $x = [\ce{H3O+}]$ : $[\ce{HA}] = c - x$ , $[\ce{A-}] = x$ (water’s own [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) neglected). 2. Solve $\dfrac{x^2}{(c - x)\,c^\circ} = K_a$ , the quadratic $x^2 + K_a c^\circ x - K_a c^\circ c = 0$ , keeping the positive root. 3. $\mathrm{pH} = -\log(x/c^\circ)$ ; check that $x$ is much larger than $10^{-7}\,\mathrm{mol}/\mathrm{L}$ . **Example 44.15 (Ethanoic acid at 0.010 mol/L0.010\,\mathrm{mol}/\mathrm{L}0.010mol/L).** $pK_a = 4.76$, so $K_a = 10^{-4.76} = 1.74 \times 10^{-5}$. Then $x^2 + 1.74 \times 10^{-5}\,x - 1.74 \times 10^{-7} = 0$ gives $x = 4.1 \times 10^{-4}\,\mathrm{mol}/\mathrm{L}$ and [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 3.39. Only $4.1 \times 10^{-4}/0.010 = 4.1\,\%$ of the acid has reacted with water; hydrochloric acid at the same concentration, [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 2.0, has reacted entirely. ![What becomes of a strong and a weak acid at the same concentration: the strong acid is entirely turned into ions, the weak acid hardly at all. Hence the pH difference, 2.0 against 3.4.](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-ka-and-pka/fig-6666879a3da3.svg) *What becomes of a strong and a [weak acid](#def-g12-ka-and-pka-strong-acid) at the same concentration: the [strong acid](#def-g12-ka-and-pka-strong-acid) is entirely turned into [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion), the [weak acid](#def-g12-ka-and-pka-strong-acid) hardly at all. Hence the [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) difference, 2.0 against 3.4.* **Safety.** ![](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-ka-and-pka/fig-f133bcfe28d0.svg) ![](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-ka-and-pka/fig-2ebfb929f83d.svg) ![](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-ka-and-pka/fig-11944cae5055.svg) ![](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-ka-and-pka/fig-b782e593bcff.svg) Concentrated methanoic acid is flammable, causes severe burns of the skin and the eyes, and its vapour is toxic. It is handled only by the teacher, under a fume hood; the dilute solutions of the exercises still need goggles. ## 44.5 Exercises **Exercise 44.1 ★.** Give the conjugate base of: $\ce{HCOOH}$, $\ce{NH4+}$, $\ce{H2O}$, $\ce{HCO3-}$; and the conjugate acid of: $\ce{NH3}$, $\ce{OH-}$, $\ce{H2O}$, $\ce{HCO3-}$. Which species are [ampholytes](#def-g12-ka-and-pka-oxonium)? **Solution of Exercise 44.1.** Conjugate bases: $\ce{HCOO-}$, $\ce{NH3}$, $\ce{OH-}$, $\ce{CO3^{2-}}$. Conjugate acids: $\ce{NH4+}$, $\ce{H2O}$, $\ce{H3O+}$, $\ce{CO2,H2O}$. [Ampholytes](#def-g12-ka-and-pka-oxonium): $\ce{H2O}$ and $\ce{HCO3-}$. **Exercise 44.2 ★.** Compute the [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) of hydrochloric acid at $0.10\,\mathrm{mol}/\mathrm{L}$, $0.0010\,\mathrm{mol}/\mathrm{L}$ and $5.0 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$. **Solution of Exercise 44.2.** [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 1.0; 3.0; $-\log(5.0 \times 10^{-3}) = 2.30$. **Exercise 44.3 ★.** A solution has [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 4.5. Compute $[\ce{H3O+}]$. A solution has [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 11.2: compute $[\ce{H3O+}]$ and $[\ce{OH-}]$ at $25\,{}^{\circ}\mathrm{C}$. **Solution of Exercise 44.3.** $10^{-4.5} = 3.2 \times 10^{-5}\,\mathrm{mol}/\mathrm{L}$. At [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 11.2: $[\ce{H3O+}] = 6.3 \times 10^{-12}\,\mathrm{mol}/\mathrm{L}$ and $[\ce{OH-}] = 1.0 \times 10^{-14}/6.3 \times 10^{-12} = 1.6 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$. **Exercise 44.4 ★.** Write the reaction with water and the expression of $K_a$ for methanoic acid and for the ammonium [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion). **Solution of Exercise 44.4.** $\ce{HCOOH + H2O <=> HCOO- + H3O+}$, $K_a = \dfrac{[\ce{HCOO-}][\ce{H3O+}]}{[\ce{HCOOH}]\,c^\circ}$; $\ce{NH4+ + H2O <=> NH3 + H3O+}$, $K_a = \dfrac{[\ce{NH3}][\ce{H3O+}]}{[\ce{NH4+}]\,c^\circ}$. **Exercise 44.5 ★.** Which is the stronger acid: methanoic acid ($pK_a = 3.74$) or ethanoic acid ($pK_a = 4.76$)? Which is the stronger base: $\ce{HCOO-}$ or $\ce{CH3COO-}$? **Solution of Exercise 44.5.** Methanoic acid (smaller $pK_a$). The stronger base is $\ce{CH3COO-}$, the base of the weaker acid. **Exercise 44.6 ★★.** A solution of a [weak acid](#def-g12-ka-and-pka-strong-acid) HA at $0.050\,\mathrm{mol}/\mathrm{L}$ has [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 3.0. Compute $[\ce{A-}]$, $[\ce{HA}]$, then $K_a$ and $pK_a$. **Solution of Exercise 44.6.** $[\ce{A-}] = [\ce{H3O+}] = 1.0 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$; $[\ce{HA}] = 0.050 - 0.001 = 0.049\,\mathrm{mol}/\mathrm{L}$; $K_a = 1.0 \times 10^{-6}/0.049 = 2.0 \times 10^{-5}$; $pK_a = 4.69$. **Exercise 44.7 ★★.** Compute the [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) of benzoic acid at $0.020\,\mathrm{mol}/\mathrm{L}$ ($pK_a = 4.20$). **Solution of Exercise 44.7.** $K_a = 6.31 \times 10^{-5}$; $x^2 + 6.31 \times 10^{-5}\,x - 1.26 \times 10^{-6} = 0$ gives $x = 1.09 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$: [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 2.96. **Exercise 44.8 ★★.** Compute the [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) of a sodium hydroxide solution at $2.0 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$. **Solution of Exercise 44.8.** $\mathrm{pH} = 14.0 + \log(2.0 \times 10^{-3}) = 14.0 - 2.70 = 11.3$. **Exercise 44.9 ★★.** Using the $pK_a$ scale, list the acids of the couples shown from the strongest to the weakest. Where would a [strong acid](#def-g12-ka-and-pka-strong-acid) such as $\ce{HCl}$ lie? **Solution of Exercise 44.9.** $\ce{HF}$, $\ce{HCOOH}$, ascorbic acid, $\ce{C6H5COOH}$, $\ce{CH3COOH}$, $\ce{CO2,H2O}$, $\ce{NH4+}$, $\ce{HCO3-}$. A [strong acid](#def-g12-ka-and-pka-strong-acid) lies far below, beyond the end of the scale: it reacts totally with water. **Exercise 44.10 ★★.** Carbon dioxide dissolved in water makes it slightly acidic (couple $\ce{CO2,H2O}$/$\ce{HCO3-}$, $pK_a = 6.37$). A sample of rain water in contact with the [air](https://one-course.com/books/chemistry/1/en/chapter/4-air-a-mixture-of-gases#def-g4-air-a-mixture-of-gases-air) is found to have [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 5.6. Compute $[\ce{H3O+}]$, and compare with pure water. **Solution of Exercise 44.10.** $[\ce{H3O+}] = 10^{-5.6} = 2.5 \times 10^{-6}\,\mathrm{mol}/\mathrm{L}$, 25 times more than in pure water ($1.0 \times 10^{-7}\,\mathrm{mol}/\mathrm{L}$). **Exercise 44.11 ★★.** Show that for the couple $\ce{NH4+}$/$\ce{NH3}$, $pK_a = pK_e - pK_b$, where $K_b = 1.8 \times 10^{-5}$ is the constant of $\ce{NH3 + H2O <=> NH4+ + OH-}$. Compute the $pK_a$. **Solution of Exercise 44.11.** $K_a K_b = \dfrac{[\ce{NH3}][\ce{H3O+}]}{[\ce{NH4+}]c^\circ} \times \dfrac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]c^\circ} = K_e$, so $pK_a = pK_e - pK_b = 14.0 - 4.74 = 9.26$. **Exercise 44.12 ★★★.** Ethanoic acid at $0.010\,\mathrm{mol}/\mathrm{L}$ has [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 3.39. Compute the [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) after a tenfold [dilution](https://one-course.com/books/chemistry/1/en/chapter/26-concentration-and-dilution#def-g10-concentration-and-dilution-dilution). Does it rise by one unit, as for a [strong acid](#def-g12-ka-and-pka-strong-acid)? Explain. **Solution of Exercise 44.12.** At $0.0010\,\mathrm{mol}/\mathrm{L}$: $x^2 + 1.74 \times 10^{-5}\,x - 1.74 \times 10^{-8} = 0$, $x = 1.24 \times 10^{-4}\,\mathrm{mol}/\mathrm{L}$, [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 3.91: a rise of 0.52, not 1. Diluted, a [weak acid](#def-g12-ka-and-pka-strong-acid) reacts in a larger proportion with water (here $12\,\%$ against $4\,\%$), which partly makes up for the [dilution](https://one-course.com/books/chemistry/1/en/chapter/26-concentration-and-dilution#def-g10-concentration-and-dilution-dilution). **Exercise 44.13 ★★★.** The hydrogencarbonate [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) is an [ampholyte](#def-g12-ka-and-pka-oxonium). Write its two reactions with water and their constants, using the $pK_a$ scale. Which is larger? What does that suggest for the [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) of a sodium hydrogencarbonate solution? **Solution of Exercise 44.13.** As an acid: $\ce{HCO3- + H2O <=> CO3^{2-} + H3O+}$, $K = 10^{-10.33}$. As a base: $\ce{HCO3- + H2O <=> H2CO3 + OH-}$ ($\ce{H2CO3}$ standing for $\ce{CO2,H2O}$), $K = K_e/K_a = 10^{-14.0+6.37} = 10^{-7.63}$. The base reaction has the larger constant: the solution is slightly basic. **Exercise 44.14 ★★★.** Hydrochloric acid at $1.0 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ is diluted a thousand times, then a thousand times again. A careless rule gives [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 6, then [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 9. Why is [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 9 impossible for a diluted acid? What [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) does the very dilute solution approach? **Solution of Exercise 44.14.** An acid, however dilute, cannot make a solution basic. When the acid’s own [oxonium ions](#def-g12-ka-and-pka-oxonium) become fewer than the $1 \times 10^{-7}\,\mathrm{mol}/\mathrm{L}$ that water provides, the water’s [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) dominate: the [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) approaches 7 from below and never passes it. **Exercise 44.15 ★★★.** Ascorbic acid (vitamin C, $pK_a = 4.04$, $M = 176.0\,\mathrm{g}/\mathrm{mol}$): a tablet of $500\,\mathrm{mg}$ is dissolved in $200\,\mathrm{mL}$ of water. Compute the [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) (consider only the first acidity). **Solution of Exercise 44.15.** $n = 0.500 / 176.0 = 2.84 \times 10^{-3}\,\mathrm{mol}$, $c = 0.0142\,\mathrm{mol}/\mathrm{L}$; $K_a = 9.12 \times 10^{-5}$; $x^2 + 9.12 \times 10^{-5}\,x - 1.30 \times 10^{-6} = 0$ gives $x = 1.09 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$: [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 2.96. ## 44.6 Problem: The Ant’s Venom **Problem 44.1.** Weekend problem — what pH does the methanoic acid of an ant’s venom give at 0.010 mol/L, and how does baking soda soothe its burn? Methanoic acid, $\ce{HCOOH}$, is found in the venom of ants. Its $pK_a$ at $25\,{}^{\circ}\mathrm{C}$ is 3.74. We study a solution at $0.010\,\mathrm{mol}/\mathrm{L}$ and compare it with hydrochloric acid. **Part I — The acid.** 1. Draw the [Lewis structure](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lewis-structure) of methanoic acid and circle the hydrogen it gives. 2. Write its couple and its reaction with water. 3. Write $K_a$ and compute its value. 4. Place the couple on the $pK_a$ scale: is methanoic acid stronger or weaker than ethanoic acid? 5. Compute the mass of methanoic acid in $1.0\,\mathrm{L}$ of the solution. **Part II — The [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph).** 6. Fill the [progress table](https://one-course.com/books/chemistry/1/en/chapter/27-the-reaction-progress-table#def-g10-reaction-progress-table-extent) per litre, with $x = [\ce{H3O+}]$ . 7. Write the equation $Q_{eq} = K_a$ . 8. Solve it. 9. Compute the [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) . 10. What fraction of the acid has reacted with water? 11. Check that the [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) coming from water can be neglected. **Part III — Comparison.** 12. What is the [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) of hydrochloric acid at $0.010\,\mathrm{mol}/\mathrm{L}$ ? 13. How many times more [oxonium ions](#def-g12-ka-and-pka-oxonium) does it contain than the methanoic acid solution? 14. Explain the difference, using the words strong and weak. **Part IV — Baking soda.** 15. Baking soda is sodium hydrogencarbonate. Write the reaction between methanoic acid and the hydrogencarbonate [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) (it gives methanoate and $\ce{CO2,H2O}$ ). 16. Show that its constant is $K = K_a(\ce{HCOOH}) / K_a(\ce{CO2,H2O})$ , and compute it. 17. Is the reaction favourable? What gas is seen? 18. Why is a base much weaker than hydroxide, such as hydrogencarbonate, preferred on the skin? 19. State the final answer: what is the [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) of a $0.010\,\mathrm{mol}/\mathrm{L}$ methanoic acid solution? **Solution of Problem 44.1.** **1.** $\ce{H-C(=O)-O-H}$: the hydrogen given is that of $\ce{O-H}$. **2.** $\ce{HCOOH}$/$\ce{HCOO-}$; $\ce{HCOOH + H2O <=> HCOO- + H3O+}$. **3.** $K_a = \dfrac{[\ce{HCOO-}][\ce{H3O+}]}{[\ce{HCOOH}]c^\circ} = 10^{-3.74} = 1.82 \times 10^{-4}$. **4.** Below ethanoic acid ($3.74 < 4.76$): stronger. **5.** $M = 46.0\,\mathrm{g}/\mathrm{mol}$: $0.010 \times 46.0 = 0.46\,\mathrm{g}$. **6.** $\ce{HCOOH}$: $0.010 - x$; $\ce{HCOO-}$: $x$; $\ce{H3O+}$: $x$. **7.** $\dfrac{x^2}{0.010 - x} = 1.82 \times 10^{-4}$. **8.** $x^2 + 1.82 \times 10^{-4}\,x - 1.82 \times 10^{-6} = 0$: $x = 1.26 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$. **9.** $\mathrm{pH} = -\log(1.26 \times 10^{-3}) = 2.90$. **10.** $1.26 \times 10^{-3}/0.010 = 12.6\,\%$. **11.** $1.26 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ is more than ten thousand times $1 \times 10^{-7}\,\mathrm{mol}/\mathrm{L}$: water’s own [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) are negligible. **12.** [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 2.0. **13.** $0.010 / 1.26 \times 10^{-3} \approx 8$ times more. **14.** Hydrochloric acid is a [strong acid](#def-g12-ka-and-pka-strong-acid), entirely turned into [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion); methanoic acid is weak: its reaction with water stops at an equilibrium where most of it remains as $\ce{HCOOH}$. **15.** $\ce{HCOOH + HCO3- -> HCOO- + CO2 + H2O}$. **16.** Multiplying top and bottom of $$ K = \frac{[\ce{HCOO-}]\,[\ce{CO2}]}{[\ce{HCOOH}]\,[\ce{HCO3-}]} $$ by $[\ce{H3O+}]$ gives $K = K_a(\ce{HCOOH}) / K_a(\ce{CO2,H2O}) = 1.82 \times 10^{-4}/4.3 \times 10^{-7} \approx 420$. **17.** Yes, $K \gg 1$; carbon dioxide bubbles out. **18.** Hydroxide is a [strong base](#def-g12-ka-and-pka-strong-acid) and would itself attack the skin; hydrogencarbonate is a [weak base](#def-g12-ka-and-pka-strong-acid) that neutralises the acid and is mild in any excess. **19.** [pH](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-ph) 2.9.