Chapter 17 Additional Aspects of Aqueous Equilibria (Part A) Often, there are many equilibria going on in an aqueous solution. So, we must determine the dominant equilibrium (i.e. the equilibrium reaction that involves major species as reactants). Major species are present at relative high concentration, whereas minor species are present at much lower concentration. Reactions between acids and bases 1. Strong Acids + Strong Base The reaction goes to completion (products are neutral). HNO 3 + KOH H 2 O + KNO 3 Net: H 3 O + + HO 2H 2 O K = 1 K w = 1.0 10 14 2. Weak Acid + Strong Base HC 2 H 3 O 2 + HO H 2 O + C 2 H 3 O 2 (reverse of K b reaction) K = 1 K b C2 H 3 O 2 = 1 = 1.8 109 5.6 10 10 The reaction goes to completion (products are basic). 3. Strong Acid + Weak Base H 3 O + + NH 3 NH 4 + + H 2 O (reverse of K a reaction) K = 1 K a NH4 + = 1 = 1.8 109 5.6 10 10 The reaction goes to completion (products are acidic). 4. Weak Acid + Weak Base May or may not go to completion. Depends on K a, K b of the acid and base mixed. 1
17.1 The Common-Ion Effect Consider a solution of acetic acid: CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO (aq) Equilibrium shifts What would happen if acetate ion, CH 3 COONa(aq), is added to the solution? (Hint: Le Châtelier.) phacid by itself < phacid and common ion The acid ionizes less (than it normally would). The same effect is obtained if we add H3O + from another source (HCl, HBr, etc.). Common-ion effect. Q. Calculate the ph of a solution containing 0.085 M nitrous acid (HNO2, Ka = 4.5 10 4 ) and 0.10 M potassium nitrite (KNO2). 17.2 Buffered Solutions A buffer solution contains a weak acid and its conjugate base (also weak); the ratio being between 1/10 and 10/1. Both HA and A are major species. A buffer solution is stable HA and A are in equilibrium with each other: HA + H 2 O H 3 O + + A 2
Buffers resist drastic ph changes when other acids or bases are added. If a strong acid is added: H 3 O + + A HA + H 2 O This uses up some A and produces some more HA. It changes the ratio of HA and A present, instead of affecting the ph directly. If a strong base is added: OH + HA H 2 O + A Converts some of the HA to A. Note: it is possible to overwhelm the buffer if you add too much strong acid or strong base. Finding the ph of a Buffer Due to the common ion effect, x s are always negligible. Initial [HA] = Equilibrium [HA] Initial [A ] = Equilibrium [A ] Check this assumption afterward (just in case). If it is not ok to make it, use successive approximations or solve the quadratic expression (your choice). For a buffer: K a = [H 3O + ][A ] [HA] Initial conc. = equilibrium conc. Since both HA and A are major species, you could use either the K a or the K b expression to solve. Important shortcut for a buffer (where x is negligible): K b = [HO ][HA] [A ] [A ] moles A = [HA] moles HA K a = [H 3 O + mol A ] mol HA K b = [OH ] mol HA mol A 3
Q. Calculate the ph of a buffer containing 0.10 mol H 2 PO 4 and 0.20 mol HPO 4 2 in 500. ml solution. Finding the ph of a Buffer Adding SA or SB to a buffer. 1. Complete reaction What s left? 2. Equilibrium. Q. If you have a buffer containing 0.0020 mol H 2 PO 4 and 0.0040 mol HPO 4 2 in 50.0 ml, what is the ph after adding 10.0 ml 0.10 M NaOH? 4
Henderson-Hasselbalch (HH) Equation It is used for buffers only. K a = [A ][H 3 O + ] [HA] ; [H 3 O + ] = K a ( [HA] [A ] ) log[h 3 O + ] = log K a ( [HA] [A ] ) log[h 3 O + ] = log K a + ( log ( [HA] [A ] )) ph = log K a + ( log ( [HA] [A ] )) ph = pk a + ( log ( [HA] [A ] )) ph = pk a + log ( [A ] [HA] ) Buffering Range A buffer will be effective when: 0.1 < [base]:[acid] < 10 Using the Henderson-Hasselbalch equation we have: Lowest ph ph = pk a + log(0.10) Highest ph ph = pk a + log(10) ph = pk a 1 ph = pk a + 1 Therefore, the effective ph range of a buffer is pka ± 1. When choosing an acid or base to make a buffer, choose one with its pka closest to the ph of the buffer. Example: Acetic Acid K a = 1.8 10 5 ; pk a = 4.74. The HC 2 H 3 O 2 /C 2 H 3 O 2 system can buffer in ph range of 3.74 5.74. If [HA] = [A ] ; ratio = 1 ; log 1 = 0 Then ph = pk a (this is a special case). If [A ] > [HA] ; more conj base: ph > pk a If [HA] > [A ]; more conj acid: ph < pk a ph of a buffer depends on 1. pka of the acid, and 2. The ratio of [base]/[acid]. 5
Making a buffer of a specific ph: 1. Choose the buffer system (conjugate acid-base pair). We want a ratio [A ]/[HA] as close to 1 as possible; so the pka of the conjugate acid should be close to the desired ph, and no more than 1 unit away (remember the ratio?) If you want a buffer with a ph = 8.00, we want a K a value close to: [H + ] = 1.0 10 8. We look at Appendix D (p. 1062). The buffer system is: HClO and ClO. HClO K a = 3.0 10 8 ; pk a = 7.52 2. Find the [A ]/[HA]. K a = [H 3O + ][ClO ] [HClO] ; K a [H 3 O + ] = [ClO ] [HClO] K a [H 3 O + ] Need 3.0 mol ClO for every 1.0 mol HClO. Could choose any amounts that fulfill this ratio. 3.0 10 8 3.0ClO = = 3.0 = 1.0 10 8 1 HClO However, if the amounts are too low, the buffer will not have a large buffer capacity (i.e. it will get used up too quickly and it won t be very resistant to ph changes). Examples: 1. Dissolve 3.0 mol NaClO and 1.0 mol HClO in 1 L H2O. 2. Mix 300 ml 0.10 M NaClO (aq) and 100 ml 0.10 M HClO. Etc. Other considerations (for the choice of a buffer) are: Solubility and Toxicity. Q. You want to make a buffer of ph = 10.00. The following compounds are available: di water, Na 2 CO 3 (s), NaHCO 3 (s), 2.00 M NH 3, NH 4 Cl s. a. Which buffer system will you use? b. If you want the least concentrated buffer component to have a concentration = 0.20 M, how will you make 500. ml of this buffer? 6
Q. c. If you want a buffer of ph 10.00 in which the least conc. component is 0.20 M, how will you make 500 ml of this buffer if available compounds are: di water, Na 2 CO 3 (s), 2.00 M HCl? 17.3 Acid Base Titrations If you know the volumes of both solutions used and the concentration of one of the solutions, you can calculate the concentration of the other solution using stoichiometry. Titrant is the substance of known concentration (added slowly); on the other hand, analyte is the substance of unknown concentration. 7
The equivalence point is the point at which the stoichiometric amount of analyte equals that of the titrant (in our case, the amount of acid equals that of base). Sometimes an indicator is used and the endpoint of the titration is found (when the indicator changes color). This point is related to the equivalence point, but is not the same. Titration Curve A plot of ph vs. amount of added titrant. The inflection point of the curve is the equivalence point of the titration. Prior to the equivalence point, the unknown solution in the flask is in excess, so the ph is closest to its ph. The ph of the equivalence point depends on the ph of the salt solution equivalence point of neutral salt, ph = 7 equivalence point of acidic salt, ph < 7 equivalence point of basic salt, ph > 7 Beyond the equivalence point, the known solution in the burette is in excess, so the ph approaches its ph 1. Titration of a SA with a SB Reaction: H 3 O + + OH 2H 2 O From the start of the titration to near the equivalence point, the ph goes up slowly. Just before (and after) the equivalence point, the ph increases rapidly. At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid. (ph=7.00) As more base is added, the increase in ph again levels off. Q. If you start with 50.0 ml 0.100 M HCl and add 40.0 ml 0.100 NaOH, calculate the ph of the resulting solution. 8
Titration of a Strong Base with a Strong Acid If the titration is run so that the acid is in the burette and the base is in the flask, the titration curve will be the reflection of the one just shown 2. Titration of a WA with a SB Unlike in the previous case, the conjugate base of the acid affects the ph when it is formed. At the equivalence point the ph is >7. Phenolphthalein is commonly used as an indicator in these titrations. At the equivalence point, all HA has been converted to A (a Weak Base), so, the solution is basic (ph>7). Halfway point: half of the initial HA has been converted to A, and half HA remains. We can estimate the pk a from the graph. [HA] = [A ] ; K a = [H 3O + ][A ] [HA] ph = pk a (at this point) = [H 3 O + ] 9
Calculating ph at different points during a titration: 1. Consider complete reactions. 2. What is left? Consider equilibria. In the first region have HA. Consider Ka equilibrium (regular WA calculation). In the buffer region, use the buffer shortcut equation. Calculations The Buffer Region Q. What is the ph of the resulting solution if 25.0 ml of 0.10 M HF is mixed with 10.0 ml of 0.15 M NaOH? K a of HF = 6.8 10 4. At the equivalence point Q. What is the ph at the equivalence point when 25.0 ml 0.10 M HF is titrated with 0.15 M NaOH? K a of HF = 6.8 10 4. 10
Beyond the Equivalence Point The excess of OH determines the ph. Q. If 25.0 ml 0.10 M HF is mixed with 25.0 ml 0.15M NaOH, what is the ph? (K a HF = 6.8 10 4 ) Titration of a Weak Acid with a Strong Base Final Notes With weaker acids, the initial ph is higher and ph changes near the equivalence point are more subtle. 3. Titration of a WB with a SA The ph at the equivalence point in these titrations is <7. Indicators Why would Phenolphthalein indicator wouldn t be a good choice for WB-SA titrations? Methyl red is the indicator of choice. ph Calculations 1. Initially there is just B. Use K b of B to calculate [OH ]. 2. Buffer region. Use buffer shortcut equation. 3. Equivalence point. Since we have BH +, we use K a and the reaction to calculate [H 3 O + ]. 4. Beyond the equivalence point, the ph is determined by the [H 3 O + ] in excess. 11
Q. Calculate the ph at the equivalence point in the titration of 25.0 ml of 0.20 M NH 3 with 0.20 M HNO 3. K b NH3 = 1.8 10 5. Choosing an Indicator Usually a small amount of indicator is used, so it doesn t affect the ph of the solution. Nevertheless, the indicator responds to the ph of the solution. The indicator for a titration must change color within the steep part of the curve. So the pk a of the indicator should be close to the ph at the equivalence point. The endpoint of a titration (the point at which the indicator changes color) ideally corresponds to the equivalence point. For SA and SB titrations, the steep part of the curve corresponds to ph ~3 11. Many different indicators would work. WA/SB titration ph at the endpoint is basic. We need an indicator with a slightly basic K b. Phenolphthalein works (ph 8-10) Methyl red doesn t (ph 4-6) WB/SA ph at the endpoint is acidic Methyl red works, but phenolpthalein doesn t. 12
Titrations of Polyprotic Acids When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation. Note: the closer the Ka s are to each other, the less distinguishable the equivalence points are More Practice Problems Q. Aniline: C 6 H 5 NH 2 K b = 4.0 10 10 dimethylamine (CH 3 ) 2 NH 2 K b = 4.0 10 4 pyridine C 5 H 5 N K b = 1.5 10 9 trimethylamine(ch 3 ) 3 NK b = 7.4 10 5 If you want a buffer at ph 5.60 a. Which is the best choice? b. What is the buffer system? c. What could you add to the WB to make the buffer? Q. You have a buffer containing H 3 BO 3 and H 2 BO 3, in which the ratio of H 2 BO 3 /H 3 BO 3 is 0.5 (or ½). K a = 5.9 10 10. a. Is this buffer acidic or basic? b. Will the ph of this buffer be greater than, less than, or equal to 9.23? 13
c. What is the ph of this buffer? d. You have 800. ml of this buffer. [H 2 BO 3 ] = 0.20 M in the buffer. What volume of 2.00 M NaOH or 2.00 M HCl is needed to change the ph to 9.50? e. What mass of H 3 BO 3 (s) or NaH 2 BO 3 (s) is needed to change the ph of he original buffer to 9.50? 14