Acids and Bases (2)

Acids and Bases 2 - Questions Link to Data Page Your feedback on these self-help problems is appreciated. Click here to send an e-mail. For on-line help on this topic, please see the following Chemcal modules: Acids and Bases (strong acids and bases; conjugate acid/base pairs; pH;self-dissociation of water) Weak Acids and Bases (weak acids; weak bases; K_a and K_b) Calculations with Weak Acids and Bases (calculations involving weak acids and bases and buffers) Acid-base Titrations (titration curves for acid/base titrations; indicators)		Shortcut to Questions Q: 1 2 3 4 5 6 7 8 9 10 11
1	Write an equation for the acid/base reaction in water solution between equimolar amounts of each of the following pairs and, using ionization constants, predict in each case whether reactants or products predominate at equilibrium (1.) HSO₄^-, SO₃^2- (2.) H₂S, H₂BO₃^- (3.) CH₃COOH, ClCH₂CO₂^- (4.) CO₃^2-, H₂CO₃ (5.) NH₄⁺, NO₂^- (6.) C₆H₅OH, HCO₃^- (7.) C₆H₅COOH, HCO₃^- (8.) Fe(OH₂)₆³⁺, F^- (9.) N₂H₅⁺, CN^- (10.) HCO₃^-, HPO₄^2-
2	Express the following H₃O⁺ concentrations in terms of pH: (1.) 1.0 x 10^-3 M (2.) 5.4 x 10^-9 M
3	Calculate [H₃O⁺] for a solution of pH 8.37.
4	Calculate the pH of a 0.10 M water solution of propanoic acid (CH₃CH₂COOH)
5	Calculate the pH of a 0.040 M aqueous solution of hydrogen cyanide
6	Calculate the pH of a 0.50 M water solution of ethylamine (CH₃CH₂NH₂)
7	At 273 K the ionic product constant of water (K_w) is 1.15 x 10^-15 M². (1.) Calculate the hydrogen ion concentration of a neutral solution at that temperature. (2.) Calculate pOH and [OH^-] for a solution of pH 4.07 at 273 K.
8	A particular water solution of formic acid is 3.2% ionised at 25 ° C. What is the pH of the solution?
9	What is the pH of a 1.0 M solution of ethanol in water?
10	Sodium (0.010 mole) is added to water (2.0 litre) (1.) What is the pH of the resulting solution? (2.) What mass of gas is given off in the reaction?
11	The average pH of beer is 4.40. If the acidity is regarded as due only to the presence of dissolved carbon dioxide from the fermentation process, calculate the concentration of the carbonic acid in the beer. (Acidity arising from the second dissociation of carbonic acid or from the dissociation of water is to be neglected).
Acids and Bases 2 (Answers)
1	The fundamental law of acid/base reactions states: "the stronger acid and base combine to give the weaker conjugates." This applies even when both acids involved are weak, but instead of complete reaction occurring as would be the case if strong species were involved, an equilibrium mixture results and the equilibrium lies towards the side with the weaker acid and base. All the parts in Q1 involve weak species and hence the use of equilibrium arrows in the following equations. (1.) HSO₄^- + SO₃^2- SO₄^2- + HSO₃^- K_A of HSO₄^- = 10^-1.99 > K_A of HSO3^- = 10^-7.20 Therefore HSO₄^- is the stronger acid and products will predominate. (2.) H₂S + H₂BO₃^- HS^- + H₃BO₃ K_A of H₂S = 10^-7.02 > K_A of H₃BO₃= 10^-9.24 Therefore H₂S is the stronger acid and products will predominate (3.) CH₃COOH + ClCH₂CO₂^- CH3CO₂^- + ClCH₂COOH K_A of ClCH₂COOH = 10^-2.86 > K_A of CH₃COOH = 10^-4.76 Therefore ClCH₂COOH is the stronger acid and reactants will predominate (4.) CO₃^2- + H₂CO₃ HCO₃^- + HCO₃ ^- K_A of H₂CO₃ = 10^-6.55 > K_A of HCO₃ ^- = 10^-10.33 Therefore H₂CO₃ is the stronger acid and reactants will predominate. (Note that there was no need to consult K_Adata as carbonic acid is a diprotic weak acid and its conjugate base, hydrogencarbonate ion, must necessarily be weaker because it has the same structure but carries a negative charge, making it more difficult for a H⁺ ion to be released.) (5.) NH₄⁺ + NO₂^- NH3 + HNO₂ K_A of HNO₂ = 10^-3.14 > K_A of NH₃ = 10^-9.24 Therefore HNO₂ is the stronger acid and reactants will predominate. (6.)C₆H₅OH + HCO₃^- C₆H₅O^- + H₂CO₃ K_A of H₂CO₃= 10^-6.55 > K_A of C₆H₅OH = 10^-9.98 Therefore H₂CO₃ is the stronger acid and reactants will predominate. (7.) C₆H₅COOH + HCO₃^- C₆H₅CO₂^- + H₂CO₃ K_A of C₆H₅COOH = 10^-4.20 > K_A of H₂CO₃ = 10^-6.55 Therefore C₆H₅COOH is the stronger acid and products will predominate. (8.) Fe(OH₂)₆³⁺ + F^- Fe(OH₂)₅(OH)²⁺ + HF K_A of Fe(OH₂)₆³⁺ = 10^-2.17 > K_A of HF = 10^-3.17 Therefore Fe(OH₂)₆³⁺ is the stronger acid and products will predominate. (9.) N₂H₅⁺ + CN^- N₂H₄+ HCN K_A of N₂H₅⁺ = 10^-7.94> K_A of HCN = 10^-9.22 Therefore N₂H₅⁺ is the stronger acid and products will predominate. (10.) HCO₃^- + HPO₄^2- H2CO₃ + H₂PO₄^- In this case, all four species ae potentially acids but both the species on the products side are diprotic acids and their conjugate bases are the reactants, so as electrically neutral H2CO₃ must be a stronger acid than its negatively charged conjugate, HCO₃^- , and single negatively charged H₂PO₄^- must be a stronger acid than its double negatively charged conjugate, HPO₄^2- , then the equilibrium lies to the reactants side.
2	pH = -log [H₃O⁺] (1.) pH = -log (1.0 x 10^-3 M) = 3.00 (2.) pH = -log (5.4 x 10^-9 M) = 8.27 Note the conversion of significant figures between concentration and pH - the number of significant figures in the concentration becomes the number of decimal places in the pH value (and vice versa), since pH is a logarithmic scale. So in the examples above, 2 significant figures of accuracy in the concentration data results in 2 decimal places of accuracy in the resultant pH values. It is usual practice to restrict the accuracy of pH values to a maximum of 2 decimal places, since measurement approximations used to calculate pH limit it's inherent accuracy.
3	[H₃O⁺] = 10^(-pH). i.e. hydrogen ion concentration is found from the inverse log of the negative pH value. ∴ Given a pH of 8.37, [H₃O⁺] = 10^-8.37 = 4.3 x 10^-9 M.
4	To calculate pH we need to know [H₃O⁺]. For a weak acid such as propanoic acid, which does not dissociate completely in water, [H₃O⁺] at equilibrium does not equal the initial concentration of the acid. So [H₃O⁺] must first be calculated: pK_A of propanoic acid = 4.87 (from data sheet). K_A = 10^-4.87 Equilibrium equation: CH₃CH₂COOH + H₂O CH₃CH₂CO₂^- + H₃O⁺. From this equation we construct the expression for K_A: K_A = [CH₃CH₂CO₂^-] [H₃O⁺] / [CH₃CH₂COOH] It is apparent from the equilibrium equation that each acid molecule that dissociates has to form one H₃O⁺ and one CH₃CH₂CO₂^- ion. Therefore [H₃O⁺] = [CH₃CH₂CO₂^-] (i.e. their concentrations must be the same). The remaining variable is [CH₃CH₂COOH]. Since propanoic acid is weak and only a very small amount is ionised in solution, we make the assumption that the equilibrium concentration of acid equals the initial concentration of acid (this allows us to avoid creating a quadratic expression, which is relatively time consuming to solve.) So in this example, we assume that the equilibrium concentration of propanoic acid = 0.10 M. Note that the validity of this assumption can be tested after calculation but for the purposes of exam questions in this chemistry course you can assume that the assumption is always valid. The worked problems presented in most textbooks do not assume the assumption is valid, and hence include checks as part of the working. Rearranging the expression for K_A: [H₃O⁺] [CH₃CH₂CO₂^-] = K_A [CH₃CH₂COOH] but since [H₃O⁺] = [CH₃CH₂CO₂^-], [H₃O⁺]² = K_A [CH₃CH₂COOH] Substituting values we obtain [H₃O⁺]² = 10^-4.87 x 0.10 = 10^-5.87 M² ∴ [H₃O⁺] = 10^-5.87/2 M = 10^-2.94 M = 1.2 x 10^-3 M Finally, from [H₃O⁺] we obtain the pH: pH = -log [H₃O⁺] = -log (1.2 x 10^-3 M) = 2.92.
5	HCN + H₂O H₃O⁺ + CN^- We find pH from [H₃O⁺], which is in turn calculated from K_A for HCN pK_A of HCN = 9.22 (from data sheet). K_A = 10^-9.22 K_A = [H₃O⁺] [CN^-] / [HCN] Since [H₃O⁺] = [CN^-] ∴ K_A = [H₃O⁺]² / [HCN] We assume [HCN] at equilibrium is equal to initial [HCN] as the amount of HCN that has dissociated is negligible compared with its initial concentration. Substituting and rearranging: K_A = 10^-9.22 = [H₃O⁺]² / 0.040 M [H₃O⁺]² = 10^-9.22 x 0.040 = 10^-10.62M² ∴ [H₃O⁺] = 10^-10.62/2 M = 10^-5.31M pH = -log [H₃O⁺] = -log ( 10^-5.31) = 5.31.
6	Ethylamine, CH₃CH₂NH₂, has the same structure as the weak base, ammonia, NH₃, except that one of the H atoms has been replaced by an ethyl group, CH₃CH₂-, but it is still a base as the central N atom has a lone pair of electrons that can bond to an H⁺ donated by an acid. The equation for the equilibrium set up by ethylamine and water is: CH₃CH₂NH₂ + H₂O CH₃CH₂NH₃⁺ + OH^- pK_A of the conjugate acid of ethylamine, the ethylammonium ion, CH₃CH₂NH₃⁺ = 10.67. Therefore pK_B of CH₃CH₂NH₂ (the conjugate base of CH₃CH₂NH₃⁺) = pK_W - pK_A(CH₃CH₂NH₃⁺) = 14 - 10.67 = 3.33. ∴ K_B for ethylamine= 10^-3.33 M. K_B = [CH₃CH₂NH₃⁺] [OH^-] / [CH₃CH₂NH₂] As there is no other source of hydroxide ions, [CH₃CH₂NH₃⁺] = [OH^-] ∴ K_B = [OH^-]² / [CH₃CH₂NH₂] We can take [CH₃CH₂NH₂] at equilibrium as equal to initial [CH₃CH₂NH₂] as the extent of hydrolysis of a weak base is negligible compared with its initial concentration. 10^-3.33 = [OH^-]² / 0.50 M [OH^-]² = 10^-3.33 x 0.50 = 10^-3.63 M² [OH^-] = 10^-3.63/2 = 10^-1.82M pOH = -log [OH^-] = 1.82 ∴ pH = 14 - pOH = 14 - 1.82 = 12.18.
7	(1.) Equation for the self-ionization of water: H₂O + H₂O H₃O⁺ + OH^- By definition, K_W = [H₃O⁺] [OH^-]. Since [H₃O⁺] = [OH^-], K_W = [H₃O⁺]² ∴ [H₃O⁺] = (K_W)^½ = (1.15 x 10^-15)^½ = 3.39 x 10^-8 M (2.) pH = 4.07. [H⁺] = 10^(-pH) = 10^-4.07 = 8.51 x 10^-5 M [OH^-] = K_W / [H⁺] = 1.15 x 10^-15 / 8.51 x 10^-5 = 1.35 x 10^-11 M ∴ pOH = -log [OH^-] = -log (1.35 x 10^-11 M) = 10.87.
8	The first step is to calculate the concentration of the HCOOH solution from K_A and the ionisation information. HCOOH + H₂O HCO₂^- + H₃O⁺ pK_A = 3.74 (from data sheet) K_A = 10 ^-3.74 M = 1.84 x 10^-4 M = [HCO₂^-] [H₃O⁺] / [HCOOH] "3.2 % ionised" implies that there are 3.2 HCO₂^-and 3.2 H₃O⁺ ions per 100 HCOOH molecules initially present. Let the initial [HCOOH] = Z M. Then [HCO₂^-] = [H₃O⁺] = 0.032 x Z M, and strictly the equilibrium [HCOOH] = Z - 0.032 Z = Z(1 - 0.032) = 0.968 Z M. However, following the procedure used previously, the equilibrium [HCOOH] can be taken as Z M. Substituting this in the expression for K_A : K_A = [HCO₂^-][H₃O⁺] / [HCOOH] = (0.032 x Z)(0.032 x Z) / Z = 1.84 x 10^-4 M ∴ (0.032 x Z)(0.032) = 1.84 x 10^-4 M Z = [HCOOH] = 1.84 x 10^-4 / (0.032 x 0.032) = 0.18 M. Now that the concentration is known, the pH is easily determined: K_A = [H₃O⁺]² / [HCOOH] ∴ [H₃O⁺] = (1.84 x 10^-4 x 0.18)^½ = 5.75 x 10^-3 M. pH = -log [H₃O⁺] = -log (5.70 x 10^-3 M) = 2.24
9	Ethanol acting as acid: CH₃CH₂OH + H₂O CH₃CH₂O^- + H₃O⁺ Water acting as acid: H₂O + H₂O OH^- + H₃O⁺ pK_A of CH₃CH₂OH is greater than 14 (from data sheet). This means that H₂O (which has a pK_A equal to 14) is a stronger acid than ethanol. The contribution of H₃O⁺ from the self ionisation of water is greater than the contribution from ethanol, and the pH of the solution is determined by the water present. As a result, the pH of the ethanol solution is 7.00 - the same as if the ethanol were not present.
10	Na + H₂O → NaOH + ½H₂ (1.) From the above equation 0.010 mole sodium will react to form 0.010 mole NaOH. Molarity = no. mol / volume = 0.010 mol / 2.0 L = 0.0050 M. NaOH is very soluble, and completely dissociates in solution, so [OH^-] = 0.0050 M. [H₃O⁺] = K_W / [OH^-] = 1.0 x 10^-14 M² / 0.0050 M = 2.0 x 10^-12 M. ∴ pH = -log (2.0 x 10^-12 M) = 11.7 (2.) Mass of H₂ gas = no. mol H₂ x molar mass H₂ = 0.0050 mol x 2.016 g mol^-1 = 0.011 g.
11	pH = 4.40, ∴ [H₃O⁺] = 10^-4.40 M. H₂CO₃ + H₂O H₃O⁺ + HCO₃^- pK_A of H₂CO₃ = 6.35. K_A = [H₃O⁺] [HCO₃^-] / [H₂CO₃] = 10^-6.35 M [H₃O⁺] = [HCO₃^-] ∴ [H₂CO₃] = [H₃O⁺]² / K_A = (3.98 x 10^-5 M)² / 4.47 x 10^-7 M = 3.5 x 10^-3 M.