Equilibrium II. Acids and Bases.

Questions and Answers for Quiz #9.






1. Define "acid" and "base" according to:

a. Arrhenius.



An acid produces H+ in water, a base produces OH-.





b. Lowry and Brønsted.



An acid is a proton donor; a base is a proton acceptor.





c. Lewis.



An acid accepts a pair of electrons in forming a covalent bond; a base donates electrons to form a covalent bond.






2a. Define "conjugate acid" and "conjugate base", and give three examples of a pair of conjugates.



Conjugates are related such that the conjugate acid has a proton mor than its conjugate base.

conjugate acid HCl NH3 HS-
conjugate base Cl- NH2- S2-







2b. Give two examples of compounds or ions which have both a conjugate acid and a conjugate base.

conjugate acid H3O+ NH4+
compound H2O NH3
conjugate base OH- NH2-







3a. Define the terms "strong" and "weak" as applied to acids and bases.



Strong and weak refer to the electrical conductivity of a solution of the acid or base. Many ions will produce a stronger current than fewer. Hence a strong acid is one the reacts well with water producing many ions; a weak acid reacts only slightly with water producing only a few ions.







3b. Describe the difference between "concentrated" and "strong", "dilute" and "weak" when applied to acids and bases.



Concentrated and dilute refer to the amount of acid or base dissolved in the water. Strong and weak refer to the number of ions produced in the reaction of the acid or base with water.








4a. Define Ka, pKa, Kb, pKb, Kw, pKw, pH, pOH.



p is the operator meaning: "take -log10 of the following quantity.





4b. What is the relationship between Ka and Kb for a conjugate acid-base pair?



For conjugates Ka * Kb = Kw





4c. What is the relationship between pH and pOH.



pH + pOH = 14 (at 25 C)








5. pH is usually considered as running from 0 to +14. Why is it difficult to talk about a pH of -6 or of +20?



The quantity having thermodynamic meaning for equilibria is the activity. Whilst the solution is dilute the concentration is a reasonable measure of the activity. This is not the case for concentrated solutions in which the activity is not directly measurable. a pH of -6 corresponds with a solution with an H3O+ activity of 1 x 106, not a dilute solution. Similarly, a pH = 20 corresponds to an OH- activity of 1 x 106, again not dilute.






6. Doubling the concentration of an HCl solution doubles the [H3O+], but does not double the pH. Explain and describe the factor by which the pH will change.



The pH scale is a logarithmic one. Doubling the [H3O+] will lead to the pH decreasing by log102 ( 0.3 units)








7. Doubling the concentration of CH3COOH (pKa = 4.8) also does not double the pH, nor, in this case does it double the [H3O+]. Explain what is happening here.



Acetic acid is a weak acid and only partially reacted with water. Increasing the concentration of the acid will increase the concentration of the [H3O+] according to the equilibrium expression:

Ka = [H3O+]2 / [HA] (approximately).

Thus doubling the acid concentration will increase the [H3O+] by approximately root 2 (1.41).






8. A 0.100 M solution of a weak acid, HA, in water has a pH of 5.5 at 298 K. What is DG for the acid reaction with water?

HA(aq) + H2O(l) H3O+(aq) + A-(aq).



If the pH = 5.5, then the [H3O+] = 3.16 x 10-6 M

and the value of K can be found:



K = [H3O+]2 / ([HA] - [H3O+]) = 1.00 x 10-10.



From DG298 = -RT ln K we find DG298 = 57.0 kJ.








9a. Prepare a graph showing the variation of the pH of a 0.1 M solution of an acid with the pKa value of the acid.



pKa = 2 * pH - log10 [HA] for small values of Ka.



If Ka is small, then the gtraph will be a straight line of slope 2.








9b. A 0.50 M solution of a weak acid has the same pH as a 0.15 M solution of HCl. What is Ka for the weak acid?



The pH of the 0.15 M HCl solution (strong acid) is .708 and [H3O+] = 0.15 M



Ka = [H3O+]2 / ([HA] - [H3O+]) = 0.0643.






9c. Cocaine is a weak base. A 5.0 x 10-3 M solution of cocaine has a pH of 10.04. What is Ka for this base?



For a base: Kb = [OH-]2 / ([B] - [OH-]) = (1.10 x 10-4)2 / (5.0 x 10-3 - 1.10 x 10-4) = 2.46 x 10-6.






9d. What is DG298 for the reaction CH3COOH(aq) + H2O H3O+(aq) + CH3COO-(aq)



Since Ka for this reaction is 1.8 x 10-5, DG298 = 27.1 kJ.






9e. A 1.00 molal solution of HF in water freezes at -l.90C. What is the value of Ka at -l.90C for HF?



The freezing point depression constant for water is 1.86 K/molal. Since the 1.00 M solution produces a solution containing 1.90/1.86 = 1.022 moles of particles, we can write:



HF + H2O H3O+ + F-



with the usual calculation for Ka = [H3O+]2 / ([HF] - [H3O+]).

and knowing that the total moles of particles is (1 - x) + x + x = 1 + x = 1.022, we find x = 0.022.



So Ka = 4.73 x 10-4.






10a. What happens to the pH of a 0.100 M HCl solution when Cl- (in the form of NaCl) is added?



No equilibrium (white lie!) to disturb and so the pH remains the same.






10b. What happens to the pH of a 0.100 M CH3COOH solution (pKa = 4.8) when CH3COO- (in the form of the sodium salt) is added?



Equilibrium here which is being disturbed by the increase in activity of the CH3COO- . This will force the equilibrium position to shift left thereby reducing the [H3O+] and increasing the pH.






10c. Why is there a difference in the two cases in 10a and 10b?



The equilibrium position is so far to the right in the first case that the chloride ion has virtually no affect on it. Not so in the second case when the equilibrium position initially lies to the left.






11. Define the term "buffer solution" and explain how it works.



A buffer solution is a solution which resists a change in the pH on addition of an acid or base to the solution.



To resist a pH change when an acid is added, the solution must contain a base.

To resist a pH change when a base is added, the solution must contain an acid.






12a. Show that the useful buffer pH range for a weak acid and its salt is pH = pKa 1.



The logarithmic version of the Ka equation can be used here:



pKa = pH - log10( [A-] / [HA]).



First: because we must work with dilute solutions, the maximum concentration for any species in solution is 1.0 M.

Second: because the solution must contain a reasonable amount of the acid and base (so that it can resist the pH change!) the minimum concentration for any species in solution is 0.1 M



If we put those limits in to give the maximum possible range we have the [A-] / [HA] ratio going from a high of 1 / 0.1 (10) to a low of 0.1 / 1 (0.1) which inside a logarithm becomes a range of +1 to -1.

Thus the range in pH is given by pKa + log10( [A-] / [HA]) or pKa 1.






12b. Show that the useful buffer pH range for a weak base and its salt is pH = 14 - (pKb 1).



Derivation is as above, but Kb must be used.






13a Show how a buffer works by comparing the change in pH which occurs when 0.1 mole of HCl(g) is added (with no change in liquid volume to

a. 1.0 L of pure water, and

b. 1.0 L of a solution containing 0.1 mole of acetic acid (pKa = 4.8) and 0.1 mole of sodium acetate.





a. 0.1 mole of HCl added to 1.0 L of pure water will give a 0.1 M solution of HCl with a pH = 1.



Range of pH change from 7 (pure water) to 1.





b. 0.1 mole HCl added to 0.1 M acetic acid and 0.1 M sodium acetate will completely react with the 0.1 moles of base (acetate ion) thus giving a solution of acetic acid (and sodium chloride) which is 0.2 M and which has a pH (from Ka and an ICE analysis) of 2.75



The original solution, with equimolar amounts of acetic acid and acetate ion has pH = pKa = 4.8.



Range of pH change from 4.8 to 2.75 (considerably less).






13b. A solution is prepared by adding 0.010 mol of solid sodium formate, Na+HCO2- to 100 mL of 0.025 M formic acid, HCOOH. Assume that no volume change occurred and calculate the pH of the resulting solution.



pH = pKa + log10( [A-] / [HA]) = 3.74 + log10 (0.10 / 0.025) = 4.34.






14. Define the terms "acid-base indicator", "end point" and "equivalence point" (making sure to distinguish between the latter two) as applied to titrations.



An acid-base indicator is a substance that has a conjugate acid base pair with different (usually intense) colours. In a titration there is a shift from one of the pair (with one colour) to the other (with a second colour).



The equivalence point is the exact point in a titration when equivalent amounts of the acid and base have been added.

The end point in a titration is where the indicator colour changes. With a suitably selected indicator, the equivalence point and the end point will be identical.






15a. Sketch the pH vs added base (0.1 M NaOH) curve for

a. the titration of 0.1 M HCl.

b. the titration of 0.1 M CH3COOH (pKa = 4.8).



15b. Sketch the pH vs added acid (0.1 M HCl) curve for

a. the titration of 0.1 M NaOH.

b. the titration of 0.1 M NH3 (pKb = 4.8)





15c. Sketch the pH vs added base (0.1 M NH3, pKb = 4.8) curve for the titration of 0.1 M CH3COOH (pKa = 4.8).





15d. Why would you never choose to titrate a weak acid with a weak base?



The pH change at the equivalence point is small and so difficult to detect.






16a. Why is the salt of a strong acid and a weak base acidic when dissolved in water? (Example NH4Cl).



The salt comprises to ions, The NH4+ ion which is the conjugate acid of the weak base NH3 and so is a weak acid, and Cl- ion, the conjugate base of the strong acid HCl and which is therefore neutral in water. Thus the salt is comprised an acidic ion and a neutral ion which leaves it acidic.



16b. Why is the salt of a strong base and a weak acid basic when dissolved in water? (Example CH3COONa).



As above but with the anion (acetate) being the conjugate base of a weak acid and therefore a weak base and the cation (sodium) being neutral in water.






17a. A warm solution of a dish-washer detergent can quickly redden the hands and lead to skin problems and so should not replace the mild dish-washing liquids sold for hand-washing of dishes. What is the main difference between the two which is causing this problem?



The difference is in the base strength of the anion in the detergent. Dish-washer detergent has a very basic anion so that the base catalysed break-down of fats and proteins left on the dishes (plus the high temperature to increase the reaction rate) can occur. For use with hands the base must be mild so that it does not catalyse the break-down of the protein (skin) of the hands.





17b. Are the foods we eat and drink usually acidic or basic? Why?



The are usually acidic/neutral.






18. What volume of 0.1050 M NaOH is required to neutralize 50.0 mL of a solution of acetic acid whose initial pH is 2.77 ? What is the pH of the resulting solution?





The reaction is: HOAc + NaOH Na+ OAc- + H2O



In order to do the stoichiometry required the initial acid concentration must be known:



Ka = [H3O+][OAc-] / [HOAc]

1.8 x 10-5 = x2 / ( a - x ) where x = [H3O+] = [OAc-] and a = initial [HOAc]



From the pH = 2.77, x = 1.70 X10-3 , and so a = 0.163 M = initial acid concentration.



Now the stoichiometry can be done:



mL NaOH = 50.0 mL HOAc*0.163 mol / L*(1 mol NaOH / 1 mol HOAc)*(1 L / 0.1050 mol)



mL NaOH = 77.6.





At the end of the titration the resulting solution contains the salt NaOAc which is a basic salt.

Kb(OAc-) must be used to find the pH:



Inital salt concentration = 50.0 ml HOAc*0.163 mol/L*(1 mol NaOAc / 1 mol HOAc)

or 8.15 mmol giving initial salt concentration = 8.15 mmol / (50 + 77.6)mL = 0.0639 M



Applying the Kb relationship:



Kb = [HOAc] * [OH-] / [OAc-] = x2 / ( b - x )



Kb = Kw / Ka = 1 x 10-14 / 1.8 x 10-5 = 5.56 x 10-10.



So

5.56 x 10-10 = x2 / (0.0639 - x )



Solving approximately x = 5.96 x 10-6, and pOH = 5.22, so pH = 8.78.






19. An unknown monoprotic acid is titrated with NaOH solution. After 12.6 mL of base is added the pH is 4.65. An additional 21.0 mL of base is required to reach the equivalence point. What is the value of Ka for the acid?



We have that Ka = [H3O+] [A-] / [HA]

and the titration reaction is: HA + NaOH NaA + H2O



Since in the Ka equation, the ratio of anion to acid concentration is required, anything directly proportional to the concentrations can be used. Since the mL of base added is directly proportional to the concentrations we can use a ratio of mL of base added.



But we have to know which number to put where.



After 12.6 mL base added, the concentration of salt will be proportional to this 12.6 value.



For the acid the concentration will be proportional to the base not yet added, or 21.0 mL.



So Ka = 2.24 x 10-5 * ( 12.6 / 21) = 1.34 x 10-5.