IB Chemistry R3.1: Proton transfer reactions worked solutions

A conjugate pair differs by exactly one \ce{H+}. \ce{HSO4-} and \ce{SO4^2-} differ by one; \ce{H2SO4} and \ce{SO4^2-} differ by two, \ce{NH4+} and \ce{NH2-} by two, and \ce{H3O+} and \ce{OH-} by two.

The concentration falls tenfold to 1.00\times10^{-3}\,\mathrm{mol\,dm^{-3}}. HCl is a strong acid, so [\ce{H+}]=1.00\times10^{-3} and \mathrm{pH}=3.00. The pH before dilution was 2.00.

Hydrochloric acid ionizes completely, so it has more ions and the higher conductivity. The pH values differ (A), equal volumes of the same alkali neutralize both (B), and excess magnesium gives the same volume of hydrogen from both (D).

Sodium methanoate contains the conjugate base of a weak acid, which takes a proton from water and gives \ce{OH-}, so the solution is alkaline. KBr is neutral; \ce{NH4NO3} and \ce{HNO3} are acidic.

Forward, \ce{H2S} loses one proton to form \ce{HS-}. In reverse, \ce{NH4+} loses one proton to form \ce{NH3}. Each is an acid in that direction.

\mathrm{pH}=-\log_{10}(2.5\times10^{-5})=4.60. The negative logarithm of hydroxide would instead give pOH.

HCl is acidic, water is neutral and NaOH is alkaline. Their pH values are approximately 1.70, 7.00 and 12.30, respectively, so the order is I, II, III.

An oxide base removes protons to form water. Copper remains \ce{Cu^2+} and needs two nitrate ions: \ce{CuO + 2HNO3 -> Cu(NO3)2 + H2O}. One mark for the correct products; one for balancing.

For this conjugate pair, K_\mathrm{b}=K_\mathrm{w}/K_\mathrm{a}=1.00\times10^{-14}/(4.0\times10^{-6})=2.5\times10^{-9}.

Only D contains a weak acid and its conjugate base. A has only a strong acid; B only a salt of a strong acid and strong base; C only strong alkali. Neither member of the buffer pair is present in A, B or C.

(a) \ce{HNO3 + NaOH -> NaNO3 + H2O}; n(\ce{HNO3})=0.0800\times0.02250=1.80\times10^{-3}\,\mathrm{mol}=n(\ce{NaOH}), so c=1.80\times10^{-3}/0.0250=0.0720\,\mathrm{mol\,dm^{-3}}. (b) [\ce{OH-}]=0.0720, [\ce{H+}]=1.00\times10^{-14}/0.0720=1.39\times10^{-13}, \mathrm{pH}=12.86. (c) The pH at equivalence is 7. The steep region covers most of pH 4 to 10, so methyl orange or phenolphthalein is acceptable.

(a) \ce{2SO2 + O2 -> 2SO3} then \ce{SO3 + H2O -> H2SO4}. (b) \ce{CaCO3 + 2H+ -> Ca^2+ + CO2 + H2O}. (c) Lime is a base: it reacts with the dissolved acid and raises the pH, and it also lowers dissolved aluminium.

(a) K_\mathrm{a}=[\ce{H+}][\ce{CH3CH2COO-}]/[\ce{CH3CH2COOH}]. (b) [\ce{H+}]=\sqrt{1.35\times10^{-5}\times0.0500}=8.22\times10^{-4}, \mathrm{pH}=3.09; assumes the ionization is small against 0.0500 (x/c=1.6\%) and that water gives negligible \ce{H+}. (c) \mathrm{p}K_\mathrm{a}=4.87, so \mathrm{p}K_\mathrm{b}=14.00-4.87=9.13. (d) The propanoate ion is the conjugate base of a weak acid: \ce{CH3CH2COO- + H2O <=> CH3CH2COOH + OH-}, so the solution is alkaline and pH is above 7.

(a) \ce{HSO3-} accepts a proton in \ce{HSO3- + H+ -> H2SO3} and donates one in \ce{HSO3- + OH- -> SO3^2- + H2O}, proving its amphiprotic behaviour. (b) Nitrogen supplies a lone pair to electron-deficient boron, so \ce{NH3} is the Lewis base. The adduct formation transfers no \ce{H+} and therefore does not demonstrate a Brønsted–Lowry base acting.

(a) In neutral water the concentrations are equal, so [\ce{H+}]=\sqrt{2.4\times10^{-14}}=1.55\times10^{-7} and neutral pH =-\log_{10}(1.55\times10^{-7})=6.81. (b) The solution has [\ce{H+}]=10^{-6.90}=1.26\times10^{-7} and [\ce{OH-}]=2.4\times10^{-14}/(1.26\times10^{-7})=1.9\times10^{-7}. Hydroxide exceeds hydrogen ions, so it is alkaline despite pH below 7.

(a) Air supplies carbon dioxide, an acidic oxide. It consumes hydroxide: \ce{CO2 + 2OH- -> CO3^2- + H2O} (hydrogencarbonate can form with further carbon dioxide). The hydroxide concentration falls, so pH falls. (b) Initial [\ce{H+}]=10^{-2.40}=3.98\times10^{-3}; after dilution it is 7.96\times10^{-5}\,\mathrm{mol\,dm^{-3}}. Thus pH =2.40+\log_{10}50=4.10. Water ionization is negligible.

(a) K_\mathrm{a}(\ce{HA})=10^{-3.20}=6.3\times10^{-4}, about 32 times the value for HB, so HA ionizes further and is stronger. (b) K_\mathrm{b}(\ce{A-})=10^{-14}/10^{-3.20}=1.6\times10^{-11}; K_\mathrm{b}(\ce{B-})=10^{-14}/(2.0\times10^{-5})=5.0\times10^{-10}. The weaker acid HB has the stronger conjugate base.

(a) \ce{HInd <=> H+ + Ind-}. Added acid raises [\ce{H+}], favouring HInd, so the mixture becomes yellow. (b) The approximate range is \mathrm{p}K_\mathrm{a}\pm1, or 5.2 to 7.2. All of this interval lies in the supplied steep region, so a small titrant volume completes the change. The equivalence point alone does not specify the whole jump width.

(a) n(\ce{NH3})=0.250\times0.0400=1.00\times10^{-2}; n(\ce{HCl})=0.200\times0.0300=6.00\times10^{-3}. After \ce{NH3 + H+ -> NH4+}: \ce{NH3} 4.00\times10^{-3}\,\mathrm{mol}, \ce{NH4+} 6.00\times10^{-3}\,\mathrm{mol}. (b) \mathrm{pOH}=4.75+\log_{10}(6.00/4.00)=4.93, \mathrm{pH}=14.00-4.93=9.07. (c) \ce{NH4+ + OH- -> NH3 + H2O}. (d) The ratio is unchanged, so the pH is about the same, but the concentrations are ten times smaller. An equal-volume sample of the diluted buffer can absorb less added acid or alkali; dilution leaves the total amounts in the whole preparation unchanged.

(a) 0.100\times20.0/25.0=0.0800\,\mathrm{mol\,dm^{-3}}. (b) 10.0\,\mathrm{cm^3} is half-equivalence, so \mathrm{p}K_\mathrm{a}=4.35 and K_\mathrm{a}=10^{-4.35}=4.5\times10^{-5}. (c) Phenolphthalein, range 8.0 to 10.0, contains the equivalence pH 8.50; methyl orange and bromothymol blue change too early. (d) [\ce{H+}]=\sqrt{4.47\times10^{-5}\times0.0800}=1.89\times10^{-3}, \mathrm{pH}=2.72.

(a) The flask initially contains an acid, and the titrant raises pH. Acid amount =0.0600\times0.0200=0.00120\,\mathrm{mol}, so equivalence volume =0.00120/0.0800=0.0150\,\mathrm{dm^3}. (b) Initial [\ce{H+}]\approx\sqrt{10^{-4.60}\times0.0600} gives pH 2.91. At 7.50\,\mathrm{cm^3}, equal acid and conjugate base give pH 4.60. At equivalence, [\ce{A-}]=0.00120/0.0350=0.0343 and K_\mathrm{b}=10^{-14}/10^{-4.60}=3.98\times10^{-10}, giving pH 8.57. Draw an ascending curve with a buffer region, a steep rise near 15.0, and a high-pH plateau. (c) Replacing the acid by equal-concentration HCl preserves the acid amount, so equivalence stays at 15.0. HCl starts at pH 1.22, has no buffer region and reaches pH 7 at equivalence. With excess NaOH, its excess amount and the total volume match, so the two curves approach the same branch.

(a) \ce{A- + H2O <=> HA + OH-}. K_\mathrm{b}=10^{-14}/10^{-4.20}=1.58\times10^{-10}; [\ce{OH-}]\approx\sqrt{1.58\times10^{-10}\times0.0400}=2.52\times10^{-6}. Thus pOH 5.60 and pH 8.40. The hydrolysis fraction is only 0.0063\%, justifying the approximation. (b) Initial pH =4.20+\log_{10}(12.0/8.00)=4.38. Added acid reacts by \ce{A- + H+ -> HA}. The amounts become 10.0 mmol each, so pH 4.20. (c) Adding 12.0 mmol HCl to a fresh original buffer consumes all 12.0 mmol A and leaves 20.0 mmol HA. With no appreciable A supplied, the buffer formula no longer applies; use weak-acid equilibrium if volume is known.

(a) The required ratio is 10^{5.30-5.00}=1.995. If the acid amount is x, then x+1.995x=30.0 mmol, so HA is 10.0158 mmol and NaA is 19.9842 mmol, or 10.0 and 20.0 mmol to three significant figures. (b) Salt mass =0.0199842\times96.0=1.92\,\mathrm{g}. Use the prescribed amounts and make the final volume to 150 cm3; volume cancels in the pH ratio. (c) \ce{HA + OH- -> A- + H2O} changes HA to 7.5158 mmol and A to 22.4842 mmol. Both remain appreciable, so pH =5.00+\log_{10}(22.4842/7.5158)=5.48. Carry the unrounded amounts between steps.

(a) \ce{CaCO3 + 2H+ -> Ca^2+ + CO2 + H2O}: two protons are used per carbonate. (b) n(\ce{CO2})=0.0545/22.7=2.40\times10^{-3}\,\mathrm{mol}=n(\ce{CaCO3}). With M(\ce{CaCO3})=100.09, the carbonate mass is 0.2403\,\mathrm{g} and its percentage is 100\times0.2403/0.320=75.1\%. (c) HCl amount =2n(\ce{CO2})=4.80\times10^{-3}\,\mathrm{mol}; the stoichiometric volume is n/c=0.0240\,\mathrm{dm^3}=24.0\,\mathrm{cm^3}. More than this is needed to ensure excess.

(a) HCl ionizes fully: pH =-\log_{10}(0.0800)=1.10. Each reagent supplies 0.0800\times0.0250=0.00200\,\mathrm{mol}, hence 0.00200 mol water forms. (b) The solution mass is 50.0 g. Q=50.0\times4.18\times0.52=108.68\,\mathrm{J}; \Delta H=-108.68/0.00200=-54.34\,\mathrm{kJ\,mol^{-1}}, or -54 to two significant figures. (c) Ethanoic acid of the same concentration initially has fewer hydrogen ions, giving a lower initial rate with identical Mg pieces at the same temperature. Both monoprotic acids contain the same total acid amount, and weak acid continues ionizing as hydrogen ions are consumed, so excess Mg gives the same final amount of hydrogen.

(a) \ce{HCOOH <=> H+ + HCOO-} gives K_\mathrm{a}=[\ce{H+}][\ce{HCOO-}]/[\ce{HCOOH}]. Rearranging, [\ce{H+}]=K_\mathrm{a}[\ce{HCOOH}]/[\ce{HCOO-}]; taking the negative logarithm gives pH =3.75+\log_{10}(9.00/6.00)=3.93. (b) \ce{HCOO- + H+ -> HCOOH} consumes 1.50 mmol methanoate and creates 1.50 mmol acid. Each becomes 7.50 mmol, so the ratio is one and pH is 3.75. (c) Equal dilution initially changes all concentrations, and a small additional dissociation restores equilibrium. For appreciable concentrations, acid and conjugate-base concentrations fall by nearly the same factor, so their ratio and pH barely change. Their concentrations per unit volume are lower, so a fixed-volume portion neutralizes less added strong acid or base.