Chapter 15 Lesson Plan Grade 12 402. The presence of a common ion decreases the dissociation. BQ1 Calculate the ph of 0.10M CH 3 COOH. Ka = 1.8 10-5. [H + ] = = ( )( ) = 1.34 10-3 M ph = 2.87 Calculate the ph of 0.10M CH 3 COOH dissolved in 0.10 M CH 3 COONa. Ka = 1.8 10-5. Given: [CH 3 COOH] =0.10M, [CH 3 COO - ] = 0.10M, K a = 1.8 10-5 CH 3 COOH(aq) CH 3 COO - (aq) + H + (aq) [initial] 0.10 0.10 - [change] -x +x +x [eqbm] 0.10 x 0.10 + x x K a = ( ) = 1.8 10-5 assume x << 0.1 [H + ] = x = 1.8 10-5 M (assumption true) ph = 4.75 BQ 1 Calculate the ph of 0.10M acetic acid solution in 0.050M sodium acetate(aq) solution. K a = 1.8 10-5. Given: [CH 3 COOH] =0.10M, [CH 3 COONa] = 0.050M, K a = 1.8 10-5 CH 3 COOH(aq) CH 3 COO - (aq) + H + (aq) [initial] 0.10 0.050 - [change] -x +x +x [eqbm] 0.10 x 0.050+x x K a = ( ) = 1.8 10-5 assume x << 0.050 x = ( )( ) = 6.0 10-3 M = [H + ] ph = 2.22 403. A buffer: is a solution that resists changes in its ph upon small additions of acid or base.sq1 SQ1. What is a buffer solution? A solution that resists change in its ph upon small addition of H + or OH -. 404. A buffer consists of: a weak acid and its conjugate base or a weak base and its conjugate acid. 405. Buffer are made in one of three ways: SQ 2 mixing a weak acid (base) with its salt mixing a weak acid (in excess) with a strong base mixing a weak base (in excess) with a strong acid SQ2 Which of the following procedures will produce a buffered solution?
i) Equal volumes of 0.3 M KOH and 1.0 M HCl solutions are mixed. ii) Equal volumes of 0.5 M KOH and 1.0M C 6 H 5 COOH solutions are mixed. iii) Equal volumes of 1.0 M C 6 H 5 COOH and 1.0 M C 6 H 5 COOK solutions are mixed. Which of the following are buffers? 0.10 M HCl and 0.10 M KCl 0.10 M HF and 0.10 M KF 0.10 M Na 2 HPO 4 and 0.10 M NaH 2 PO 4 Given 1:1 molar ratio of: NH 3 / NH 4 Cl - which solution has the lowest ph? H 3 PO 4 / NaH 2 PO 4 - which solution is almost neutral? HCl / NaCl - which solution is a buffer at ph > 8? NaOH / NH 3 - which solution is a buffer at ph < 6? NH 3 / CH 3 COOH 406. Buffers work: because they have species that neutralized added H + or OH -. SQ3 HF/F - Buffer: If H + added: H + + F - HF (H + neutralized) If OH - added: HF + OH - F - + H 2 O (OH - neutralized) NH 3 /NH + 4 Buffer: If H + added: H + + + NH 3 NH 4 (H + neutralized) If OH - added: NH + 4 + OH - NH 3 + H 2 O (OH - neutralized) SQ 3 Explain how a CH 3 COOH / CH 3 COO - buffer maintains ph relatively the same upon additions of a few drops of HCl? Of NaOH? When HCl(aq) is added, the added H + will be neutralized by the acetate ions present in the buffer solution. CH 3 COO - (aq) + H + (aq) CH 3 COOH(aq) When NaOH(aq) is added, the added OH - will be neutralized by the acetic acid molecules present in the buffer solution. CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O(l) 407. ph of buffer solutions: can be found using the Henderson Hasselbalch equation. SQ4, BQ2, ex 31, 33 SQ 4 What is the Henderson - Hasselbalch equation? ph = pka + log BQ 2 Calculate the ph of a buffer solution which is 0.2M NaF and 0.3M HF. Given: [NaF] = 0.2M, [HF] = 0.3M, K a = 7.2 10-4 It is a buffer solution, therefore ph = pka + log = -log7.2 10-4 + log = 3.14 +( 0.18) = 2.96 How can we prepare a buffer with ph = 7 using only solutions of 0.10 M CH 3 COOH and CH 3 COONa? Ka = 1.8 10-5. ph = pka + log 7 = 4.75 + log = 180 Use much more of the acetate ion than the acetic acid
408. If is constant, ph of the buffer is unchanged: dilution has no effect on ph of buffer. BQ3, 4 BQ 3 Calculate the ph of a buffer solution which is 0.4M NaF and 0.6M HF. (Compare your result with the one you got in question 2 above!) ph = 2.96, the ph remains the same as the change in the concentrations did not change the ratio of [base] to [acid]. BQ 4 A buffer solution is prepared which is 0.2M NaF and 0.3M HF. 100ml of water were added to 100 ml of the above solution. Calculate the new ph. ph = 2.96, the addition of water dilutes both the acid and the base. The ratio of [base] to [acid] does not change, so the ph does not change. 409. Buffering capacity: measures the ability of a buffered solution to absorb H + or OH - without significant change in its ph. The larger the concentrations of the weak acid (base) and its conjugate base (acid), the higher the capacity. A ratio of important than high concentrations). SQ5, 6 = 1, gives the best buffering capacity. (The ratio being equal to 1 is more SQ5 What is meant by the buffering capacity of a buffer solution? It is the ability of a buffered solution to absorb H + or OH - ions without a significant change in ph. SQ6 Which ratio of [conjugate base] / [weak acid] would produce the most efficient buffer? = 1 410. When choosing an acid to make a buffer of a given ph, the pka of the acid must be as close as possible to the required ph. SQ7, BQ6 SQ7 You wish to create a buffer solution with a ph of 5.00. Which of the following would be the best choice to prepare the required buffer? i) HCl ii) H 3 AsO 4 K a =5.5 x 10-3 iii) CH 3 COOH K a = 1.8 x 10-5 iv) HCN K a = 4.9 x 10-10 i) HCl Ka very large pka = very negative ii) H 3 AsO 4 K a =5.5 x 10-3 pka = 2.3 iii) CH 3 COOH K a = 1.8 x 10-5 pka = 4.7 iv) HCN K a = 4.9 x 10-10 pka = 9.3 Best acid to choose is CH 3 COOH K a = 1.8 x 10-5 pka = 4.7 BQ 6 A buffer solution of ph = 5 is to be prepared using CH 3 COONa and CH 3 COOH. What should be the molar ratio [CH 3 COO - ] to [CH 3 COOH] to achieve this? Given: ph = 5, K a = 1.8 10-5 RTF: ph = pk a + log = - log1.8 10-5 + log = 5
log = 5 4.74 = 0.26 = 1.82 A buffer made using equimolar amounts of NaHSO 3 and Na 2 SO 3 would be most effective between the ph ranges (Ka H 2 SO 3 = 1 10-2, Ka HSO 3 - = 1 10-7) : 8 10 6 8 4 6 2 4 0.5 2 411. Steps to find ph of solution when a strong base/strong acid is added to a buffer. BQ 5 BQ 5 A buffer solution is prepared which is 0.10M CH 3 COONa and 0.20M CH 3 COOH. 10cm 3 of 0.20M NaOH(aq) were added to 100cm 3 of the above solution. Find the new ph. Given: [CH 3 COOH] = 0.20M, [CH 3 COO - ] = 0.10M, V buffer = 100cm 3, V NaOH = 10cm 3 [NaOH] = 0.20M, K a = 1.8 10-5 Step 1 Find number of moles of each species before mixing n of OH - = (0.20)(10) = 2.0 mmoles n of CH 3 COOH = (0.20)(100) = 20mmoles n of CH 3 COO - = (0.10)(100) = 10mmoles Step 2 Find number of moles left after mixing Write the reaction occurring: CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O(l) Initially 20 mmoles 2.0 mmoles 10 mmoles Finally 18 mmoles ---- 12 mmoles The resulting solution contains a weak acid with its conjugate base, a buffer. Step 3 Use Hendersen Hasselbalch equation to find ph ph = pk a + log = -log1.8 10-5 + log = 4.56 412. Steps to find ph of a solution formed by mixing: Weak acid (in excess) with strong base. BQ7 BQ7 A solution is prepared by mixing 70ml, 0.2M HF and 30ml, 0.1M NaOH. Calculate the ph of the solution. Given: V HF = 70ml, [HF] = 0.2M, V NaOH = 30ml, [NaOH] = 0.1M, K a = 7.25 10-4 n HF = (70)(0.2) = 14mmoles n NaOH = (30)(0.1) = 3mmoles HF(aq) + OH - (aq) F - (aq) + H 2 O(aq) Initially 14 3 - Finally 11-3 ph = pk a + log = - log7.25 10-4 + log = 3.14 +( 0.56) = 2.58 413. Steps to find ph of a solution formed by mixing: Weak basic salt (in excess) with strong acid. BQ 9 BQ 9 A solution is prepared by mixing 60ml, 0.2M NaF and 40ml, 0.1M HCl. Calculate the ph of the solution. Given: V NaF = 60ml, [NaF] = 0.2M, V HCl = 40ml, [HF] = 0.1M, K a HF = 7.25 10-4
n of F - = (60)(0.2) = 12mmoles n HCl = (40)(0.1) = 4mmoles F - (aq) + H + (aq) HF(aq) Initially 12 4 - Finally 8-4 ph = pk a + log = - log7.25 10-4 + log = 3.44 414. Steps to find ph of a solution formed by mixing: Weak base (in excess) with strong acid BQ10 BQ 10 A solution is prepared by mixing 20ml, 3M NH 3 and 80ml, 0.2M HCl. Calculate the ph of the solution. Given: V = 20ml, [NH 3 ] = 3M, V HCl = 80ml, [HCl] = 0.2M, K b NH 3 = 1.8 10-5 NH 3 n of NH 3 = (20)(3) = 60mmoles n HCl = (80)(0.2) = 16mmoles NH 3 (aq) + H + (aq) NH + 4 (aq) Initially 60 16 - Finally 44-16 K a = = 5.56 10-10 ph = pka + log = - log5.56 10-10 + log =9.25 + 0.44 = 9.69 415. Steps to find ph of a solution formed by mixing: Acid salt (in excess) with strong base. BQ 8 BQ 8 A solution is prepared by mixing 80ml, 0.3M NH 4 Cl and 20ml, 0.2M NaOH. Calculate the ph of the solution. Given: V NH4Cl = 80ml, [NH 4 Cl] = 0.3M, V NaOH = 20ml, [NaOH] = 0.2M, K b of NH 3 = 1.8 10-5 Calculate the number of moles: n of NH 4 + = (80)(0.3) = 24mmoles n NaOH = (20)(0.2) = 4mmoles NH + 4 (aq) + OH - (aq) NH 3 (aq) + H 2 O(aq) Initially 24 4 - Finally 20-4 Ka = = 5.56 10-10 ph = pk a + log = - log(5.56 10-10 ) + log =9.25 0.70 = 8.55 416. Titration: a technique in which one solution (titrant) is used to analyze another (analyte). 417. Equivalence point: is the point in the titration when enough titrant has been added to react exactly with the substance being titrated, analyte. SQ 8 SQ8 What is meant by equivalence or stoichiometric point in titration? Equivalence or stoichiometric point is is the point in the titration when enough titrant has been addded to react exactly with the substance in solution being titrated.
418. End point: the point in the titration when the indicator changes colour. SQ 9 SQ9 What is meant by end point in titration? End point is the point in the titration at which the indicator changes colour. 419. At the equivalence point of acid-base titrations: moles of H + = moles of OH -. BQ 11, 14 BQ 11 a) 25cm 3 of a certain NaOH solution were needed to neutralize 20cm 3 of a 0.20M HCl solution. Calculate the concentration of the base. Given: [HCl] = 0.20M, V HCl = 20cm 3, V NaOH = 25cm 3 RTF: [NaOH] At the equivalence point: number of moles of H + = number of moles of OH - [H + ] = [OH - ] (0.20)(20) = [OH - ](25) [OH - ] = 0.16M b) What volume of 0.5M HCl is needed to neutralise 40cm 3 of a 0.02M Ca(OH) 2 solution? Given: [HCl] = 0.50M, ( ) = 40cm3, [Ca(OH) 2 ] = 0.02M RTF: [HCl] At the equivalence point: number of moles of H + = number of moles of OH - # of moles of OH - = 2 # of moles of Ca(OH) 2 = 2(0.02) = 0.04M [H + ] = [OH - ] (0.50) = (0.04)(40) = 3.2cm 3 Another way to solve this type of exercise is to write the reacting ratio: Ca(OH) 2 (aq) + 2HCl(aq) CaCl 2 (aq) + 2H 2 O(l) 1 mole 2 moles ( ) n HCl n HCl = 2 ( ) (0.50)( ) = 2(0.02)(40) = 3.2cm 3 BQ 14 0.20g of a monoprotic acid was dissolved in 25 cm 3 solution. The solution was titrated with 0.125M NaOH solution. 20 cm 3 of the base was needed to reach the equivalence point. What is the molar mass of the acid? Given: m acid = 0.20g, V acid = 25 cm 3, [NaOH] = 0.125 M, V base = 20 cm 3 RTF: Molar mass of acid Since monoprotic acid, at the equivalence point, n acid = n base = [base]v base = (20 10-3 )(0.125) M acid = 80 g/mol 420. Strong acid-strong base titration. BQ12, ex. 57
BQ 12 a. 50 cm 3 of 0.50 M HNO 3 solution were titrated with 0.50 M NaOH solution. Find the ph of the solution when no base has been added yet. Given: = 50 ml, [HNO 3 ] = 0.50M, [NaOH] = 0.50M, V NaOH = 0ml HNO 3 is a strong acid, it dissociates completely. [H + ] = [HNO 3 ] = 0.50M ph = -log0.50 = 0.30 b. 50 cm 3 of 0.50 M HNO 3 solution were titrated with 0.50M NaOH solution. Find the ph of the solution when 10cm 3 of base are added. Given: = 50 ml, [HNO 3 ] = 0.50M, [NaOH] = 0.50M, V NaOH = 10ml = (50)(0.50) = 25mmoles n NaOH = (10)(0.50) = 5mmoles n of acid in excess = 25 5 = 20 mmoles [H + ] = [HNO 3 ] excess = = 0.33 M ph = -log0.33 = 0.48 c. 50 cm 3 of 0.50 M HNO 3 solution were titrated with 0.50 M NaOH solution. Find the ph of the solution when 50cm 3 of base are added. Given: = 50 ml, [HNO 3 ] = 0.50M, [NaOH] = 0.50M, V NaOH = 50ml = (50)(0.50) = 25mmoles n NaOH = (50)(0.50) = 25mmoles n of acid in excess = 25 25 = 0 mmoles Neither H + nor OH - is in excess, solution is neutral ph = 7 d. 50 cm 3 of 0.50 M HNO 3 solution were titrated with 0.50 M NaOH solution. Find the ph of the solution when 60cm 3 of base are added. Given: = 50 ml, [HNO 3 ] = 0.50M, [NaOH] = 0.50M, V NaOH = 60ml = (50)(0.50) = 25mmoles n NaOH = (60)(0.50) = 30mmoles n of base in excess = 30 25 = 5 mmoles NaOH is a strong base, it dissociates completely. noh - = n NaOH = 5mmoles [OH - ] excess = = 0.045 M poh = -log0.045= 1.34 ph = 14 poh = 14 1.34 = 12.7 e. 50 cm 3 of 0.50 M HNO 3 solution were titrated with 0.50 M NaOH solution. Draw a titration curve for the change of ph versus volume of base added. In your curve the following points have to be clearly
indicated: The ph before the titration starts and at the equivalence point. The volume of base needed for the equivalence point should be also shown. V NaOH (cm 3 ) ph 0 0.30 20 0.70 30 0.90 40 1.26 50 7.0 60 12.7 ph 13 12 11 10 9 8 7 equivalence point ph = 7 6 5 4 3 2 1 10 20 30 40 50 60 70 80 90 100 Volume of NaOH (cm 3 ) 421. Strong base - weak acid titration. BQ13, ex. 61 BQ 13 a. 50 cm 3 of 0.10 M CH 3 COOH solution were titrated with 0.10 M KOH solution. Find the ph of the solution when no base has been added yet. Ka = 1.80 10-5 Given: = 50 ml, [CH 3 COOH] = 0.10M, [KOH] = 0.10M, V KOH = 0ml, Ka = 1.80 10-5 CH 3 COOH is a weak acid, it dissociates partially: CH 3 COOH(aq) CH 3 COO - (aq) + H + (aq) [initially] 0.10M - - [cange] -x +x +x [eqbm] 0.10 x x x Ka = = 1.80 10-5 assume x << 0.10
[H + ] = x = ( )( = 1.34 10-3 M (assumption valid) ph = -log1.34 10-3 = 2.87 b. 50 cm 3 of 0.10 M CH 3 COOH solution were titrated with 0.10 M KOH solution. Find the ph of the solution when 10 ml of base has been added. Ka = 1.80 10-5 Given: = 50 ml, [CH 3 COOH] = 0.10M, [KOH] = 0.10M, V KOH = 10ml, Ka = 1.80 10-5 = (50)(0.10) = 5mmoles n KOH = (10)(0.10) = 1mmole CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O(l) Initially 5mmoles 1mmole - After reaction 4mmoles - 1mmole After the reaction we end up with a solution containing a weak acid and its conjugate base buffer solution. ph = pk a + log = 4.74 0.60 = 4.14 c. 50 cm 3 of 0.1 M CH 3 COOH solution were titrated with 0.1 M KOH solution. Find the ph of the solution when 25cm 3 base are added. (What is the significance of this stage of titration?) Given: = 50 ml, [CH 3 COOH] = 0.10M, [KOH] = 0.10M, V KOH = 25ml, Ka = 1.80 10-5 = (50)(0.10) = 5mmoles n KOH = (25)(0.10) = 2.5mmoles CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O(l) Initially 5mmoles 2.5mmole - After reaction 2.5mmoles - 2.5mmoles After the reaction we end up with a solution containing a weak acid and its conjugate base buffer solution. ph = pk a + log = 4.74 This point is called the half-way equivalence point. The buffer is at its best capacity and the ph = pk a d. 50 cm 3 of 0.1 M CH 3 COOH solution were titrated with 0.1 M KOH solution. Find the ph of the solution when 50cm 3 base are added. (What is the significance of this stage of titration?)
Given: = 50 ml, [CH 3 COOH] = 0.10M, [KOH] = 0.10M, V KOH = 50ml, Ka = 1.80 10-5 = (50)(0.10) = 5 mmoles n KOH = (50)(0.10) = 5 mmoles CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O(l) Initially 5mmoles 5mmole - After - - 5mmoles reaction [CH 3 COO - ] = = 0.050M After the reaction we end up with a solution containing a weak base CH 3 COO - (aq). CH 3 COO - (aq) + H 2 O(l) CH 3 COOH(aq)+ OH - (aq) [initially] 0.050M - - [change] -x +x +x [eqbm 0.050 x x x Kb = = = 5.56 10-10 assume x << 0.050 [OH - ] = x = ( )( ) = 5.27 10-6 M (assumption valid) poh = 5.28 ph = 14 poh = 14 5.28 = 8.72 e. 50 cm 3 of 0.1 M CH 3 COOH solution were titrated with 0.1 M KOH solution. Find the ph of the solution when 60cm 3 base are added. Given: = 50 ml, [CH 3 COOH] = 0.10M, [KOH] = 0.10M, V KOH = 60ml, Ka = 1.80 10-5 = (50)(0.10) = 5mmoles n KOH = (60)(0.10) = 6mmoles n of base in excess = 6 5 = 1 mmole KOH is a strong base, it dissociates completely. noh - = n KOH = 1mmole [OH - ] excess = = 9.09 10-3 M poh = -log 9.09 10-3 = 2.05 ph = 14 poh = 14 2.04 = 11.95 f. 50 cm 3 of 0.1 M CH 3 COOH solution were titrated with 0.1 M KOH solution. Draw a titration curve for the change of ph versus volume of base added. In your curve the following points have to be clearly indicated: The ph before the titration starts and at the equivalence point. The volume of base needed for the equivalence point should be also shown.
Volume of base added ph 0 cm 3 2.87 10 cm 3 4.14 20 cm 3 4.56 30 cm 3 4.92 40 cm 3 5.34 49.95 cm 3 6.73 50 cm 3 8.72 50.05 cm 3 9.70 60 cm 3 11.95 70 cm 3 12.7 ph 13 12 11 10 9 8 7 6 5 4 3 2 1 10 20 30 40 50 60 70 Volume of KOH (cm 3 ) 422. Half equivalence point: is when half the acid/base is neutralized. At that point, n HA = ph = pka Buffer is at its best capacity. SQ 12 SQ12 In the titration of weak acid with strong base what is meant by "the half way point"? What can you say about the buffering action of the resulting solution? How does the ph at the half-equivalence point compare with the value of Ka? The half equivalence point is the point in the titration when half the starting material has been titrated. At the half-equivalence point, ph = pka, and the buffer is at its best buffering capacity. 423. Comparison between titration curves. SQ11
SQ 11 base? How is the titration curve for a weak acid-strong base different from that of a strong acid-strong Strong Acid strong base Titration Weak Acid strong base Titration No small initial rise A small initial rise Its vertical section is longer Its vertical section is shorter Its rise is sharp Its rise is not sharp Before equivalence point the ph steadily increases Before the equivalence point, there is an almost horizontal section where the solution behaves as a buffer ph at equivalence point = 7 ph at equivalence point is greater than 7 424. The effect of acid strength and quantity on its titration with a strong base. BQ15 Acid strength Acid quantity Volume of base needed No effect The greater the quantity ph at the equivalence point The stronger the acid the closer the ph to 7 No effect Shape of ph curve The stronger the acid the sharper the curves and the longer the vertical section Same general shape List, other than water, all the species present in a solution of weak acid HA(aq) in decreasing concentration. HA > H + = A - > OH - Correct procedures for a titration include: Draining the pipette by touching the tip to the side of the container used for titration. Rinsing the buret with distilled water just before filling it with the liquid to be titrated. Swirling the solution frequently during the titration. A, C This graph was obtained when 0.10M acid was titrated with NaOH. How much is pka? ph = pka at half equivalvence point. Equivalence point is at V = 50 cm 3 half - equivalence point is at V = 25 cm 3. ph at that point is around 4. A student titrated a weighed sample of a solid acid (HA) with standard aqueous base to find its molar mass. What effect will each of the following have on the determined molar mass?
Adding more water than is needed to dissolve the acid. No effect Adding some base beyond the equivalence point. Obtained molar mass too small Failure to rinse all the acid from the weighing paper into the titration vessel. Obtained molar mass too large A 3.00g sample of a weak acid HA was titrated with 1.0 M NaOH. The ph titration curve obtained is shown below. Find pka and the molar mass of the acid. pka around 6 n acid = n base m M = CV M = m CV 3 00 (1 0)(20 10 3 ). M = 150 g/mole An unknown weak acid HA was dissolved and titrated with NaOH. Titration curve obtained shown, find pka. ph = pka = around 4 BQ 15 25 ml of 0.20M HA was titrated with 0.20M NaOH. The ph at the equivalence point was 8.00. What is K a of HA? (Are the volumes necessary in this problem? Why?) Given: V acid = 25 cm 3, [acid] = 0.20M, [NaOH] = 0.20 M RTF: Ka At equivalence point, we have a solution of A -. HA(aq) + OH - (aq) A - (aq) + H 2 O(l) Initially 5mmoles 5mmole - After reaction - - 5mmoles
[A - ] = ( )( ) = 0.10M A - is a base, it reacts with water setting up the following equilibrium: A - (aq) + H 2 O(l) HA(aq) + OH - (aq) [initially] 0.10M - - [change] -x +x +x [eqbm 0.10 x x x ph is given to be 8.00 at equivalence point poh = 6 At equivalence point, [OH - ] = 1.0 10-6 M [HA] = 1.0 10-6 M K b = = ( ) = 1.0 10-11 K a = = 1.0 10-3 425. Acid base indicator: is a substance, usually a weak organic acid, that changes color within a fairly narrow range of ph values because the acid and its conjugate base have different colors. SQ 14 SQ 14. What is an acid-base indicator? An acid base indicator is a substance, usually an organic acid,that changes color within a fairly narrow range of ph values because the acid and its conjugate base are of two different colors. 426. Indicators you must know: Indicator Color in Acid Color in Base Changes color at ph Phenolphthalein Colorless Pink 9 Methyl orange Red Yellow 4 427. Indicator changes color over the range: its pka 1 Suppose an indicator HIn has a Ka = 1.0 10-5. HIn is yellow and In - is red. If placed in an acidic solution, it will exist as HIn solution s color is yellow. As a base is added, HIn In -. A color change occurs when [In - ] = [HIn]. Since both HIn and In - are present, then the indicator acts as a buffer: ph = pka + log. Solution becomes orange when ph = -log 1.0 10-5 + log = 5.0-1 = 4.0 If placed in a basic solution, it will exist as In - solution s color is red. As a acid is added, In - HIn. A color change occurs when [HIn] = [In - ]. Solution becomes orange when ph = -log 1.0 10-5 + log = 5.0 + 1 = 6.0 Summary: When = 0.1 we see color of HIn When we see color of In - When 0.1 = < we see intermediate color
Useful ph of indicator is pka 1 When ph > pka + 1 we see color of In - When ph < pka 1 we see color of HIn When pka + 1 > ph > pka 1 we see color of intermediate 428. In choosing an indicator for a titration: we want the color change to occur approximately at the ph of the equivalence point. Since ph changes rapidly around the equivalence point, we need not be exact. This is especially true for strong acid strong base titrations. When a strong acid is titrated with a strong base using phenolphthalein, the color changes suddenly at the end point. The color can be switched back and forth by addition of only a single drop of acid or base. What is the reason for the sudden color change? A large change in ph occurs near the enpoint of a titration 429. In choosing an indicator for a titration:its useful ph range must lie completely within the vertical part of the ph titration curve. 430. For a strong acid strong base titration there is a wider choice of indicators than for a strong acid weak base or weak base strong acid titration. SQ15, MCBQ 1-8, BQ 16, 18 SQ 15 In a particular titration experiment the ph of the solution changed from 4.0, one drop before the equivalence point to 10, one drop after the equivalence point. Will any of these two indicators, with their range of colour change be suitable for the above titration? HIn' : 3.8 to 5.4; HIn" : 8 to 9.8 The best indicator is one whose ph range change lies completely within the ph rise. HIn" has a ph color change between 8 to 9.8, which lies completely between 4.0 and 10. Therefore, HIn is the better indicator to choose for this titration. BQ 16 A solution of 0.100 M HCl and a solution of 0.100 M NaOH are prepared. A 40.0 ml sample of one of the solutions is added to a beaker and then titrated with the other solution. A ph electrode is used to obtain the data that are plotted in the titration curve shown below. (b)
(a) Identify the solution that was initially added to the beaker. Explain your reasoning. The solution in the beaker was the 0.100 M HCl because the initial ph was 1 (the ph of 0.100 M HCl). (b) On the titration curve above, circle the point that corresponds to the equivalence point. The point with coordinates (40.0, 7) is circled. (c) At the equivalence point, how many moles of titrant have been added? Given: [NaOH] = 0.100M RTF: n n NaOH = (0.100)(40.0 10-3 ) = 0.00400 mol NaOH (d) The same titration is to be performed again, this time using an indicator. Use the information in the table below to select the best indicator for the titration. Explain your choice. Indicator ph range of color change Crysatl violet 0 1.8 Bromocresol Purple 5.2 6.9 Thymophthalein 9.5 10.5 Bromocresol purple would be best because its color change will occur closest to the equivalence point (at equivalence point ph changes from about 4 to 10). (e) What is the difference between the equivalence point of a titration and the end point of a titration? The equivalence point in a titration occurs when the number of moles of titrant added is exactly sufficient to react completely with the number of moles of the titrated species present in the sample being titrated. The end point of a titration is the point in a titration at which the indicator undergoes its color change. (f) On the grid provided below, sketch the titration curve that would result if the solutions in the beaker and buret were reversed (i.e., if 40.0 ml of the solution used in the buret in the previous titration were titrated with the solution that was in the beaker). ph
E ph 8.7 D C B A Volume of base added The diagram represents the ph titration curve of a weak monoprotic acid with a 0.100 M NaOH. Use it to answer the following questions. 1. The ph at this point is less than 3. A 2. The ph at this point is greater than 8 and less than 10. D 3. The ph at this point could be used to determine the acid dissociation constant. B 4. What part of the curve corresponds to the optimum buffer action? B 5. At this point, the solution is buffered. B 6. Of the points shown on the graph, this is the point when the solution is most basic. E 7. Of the points shown on the graph, this is the point is the equivalence point of the titration. D 8. Which of the following indicator is the best choice for the titration? [-A-] Methyl orange 3.2 4.4 [-D-] Phenol phthalein 8.2 10.0 [-B-] Methyl red 4.8 6.6 [-E-] Alizarin 11.0 12.4 [-C-] Bromothymol blue 6.1 7.6 BQ 18 CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) Acetic acid dissociates in water as shown in the equation above. A 25.0ml sample of aqueous solution of 0.100M acetic acid is titrated using standardized 0.100M NaOH. a) After addition of 12.5 ml of 0.100M NaOH, the ph of the resulting solution is 4.74. Calculate each of the following. (i) [H + ] in solution Given: = 25 ml, [CH 3 COOH] = 0.100M, [NaOH] = 0.100M, V NaOH = 12.5ml, Ka = 1.80 10-5, ph = 4.74 RTF: [H + ] This point is called the half-way equivalence point. At this point the ph = pka. [H + ] = K a = 1.80 10-5 M (ii) [OH - ] in the solution Given: [H + ] = 1.80 10-5 M RTF: [OH - ] [OH - ] = = 5.56 10-10 M
(iii) The number of moles of NaOH added Given: [NaOH] = 0.100M, V NaOH = 12.5ml RTF: n NaOH n NaOH = (0.100)(12.5 10-3 ) = 1.25 10-3 mole (iv) The number of mole of CH 3 COO - in the solution. Given: = 25 ml, [CH 3 COOH] = 0.100M, [NaOH] = 0.100M, V NaOH = 12.5ml, Ka = 1.80 10-5 RTF: [CH 3 COO - ] = (25)(0.10) = 2.5mmoles n NaOH = (12.5)(0.10) = 1.25mmoles CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O(l) Initially 2.5mmoles 1.25mmol - e After reaction 1.25mmoles - 1.25mmoles v) The number of moles of CH 3 COOH present the solution Given: = 25 ml, [CH 3 COOH] = 0.100M, [NaOH] = 0.100M, V NaOH = 12.5ml, Ka = 1.80 10-5 RTF: [CH 3 COOH] = (25)(0.10) = 2.5mmoles n NaOH = (12.5)(0.10) = 1.25mmoles CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O(l) Initially 2.5mmoles 1.25mmole - After reaction 1.25mmoles - 1.25mmoles b) State whether the solution at the equilibrium of the titration is acidic, basic, or neutral. Explain your reasoning. At the equivalence point, we have a solution of CH 3 COO -. This ion dissociates in water as follows: CH 3 COO - (aq) + H 2 O(l) CH 3 COOH(aq) + OH - (aq) Since OH - ion are formed then the solution is basic. In a different experiment, 0.118 g sample of a mixture of solid CH 3 COOH and solid NaCl is dissolved in water and titrated with 0.100M NaOH. The equivalence point was reached when 10.00 ml of the base solution is added. c) Calculate the following: (i) The mass in grams of acetic acid solid in the mixture Given: [NaOH] = 0.100M, V NaOH = 10.00ml RTF: m of CH 3 COOH At the equivalence point, n NaOH = n acid since the base is monobasic and the acid is monoprotic n of acid = (0.100)(10.00) = 1.00 mmole m = (1.00 10-3 )(60) = 0.0600g (ii) The mass percentage of acetic acid in the solid Given: m of acid = 0.0600g, m total = 0.118g RTF: mass % of acid m % = = 50.8%
d) Calculate the ph at the equivalence point of the titration of 25.0 ml 0.10M CH 3 COOH with 0.10 NaOH solutions. Given: = 25.0 ml, [CH 3 COOH] = 0.10M, [NaOH] = 0.10M, V KOH = 25.0ml, Ka = 1.80 10-5 = (25.0)(0.10) = 2.5 mmoles n KOH = (25.0)(0.10) = 2.5 mmoles CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O(l) Initially 2.5mmoles 2.5mmole - After reaction - - 2.5mmoles [CH 3 COO - ] = = 0.05M After the reaction we end up with a solution containing a weak base CH 3 COO - (aq). CH 3 COO - (aq) + H 2 O(l) CH 3 COOH(aq)+ OH - (aq) [initially] 0.05M - - [change] -x +x +x [eqbm 0.05 x x x K b = = = 5.56 10-10 assume x << 0.050 [OH - ] = x = ( )( ) = 5.27 10-6 M (assumption valid) poh = 5.28 ph = 14 poh = 14 5.28 = 8.72 e) The pka values for several indicators are given below. Which of the indicators listed is most suitable for this titration? Justify your answer. Indicator pka Erythrosine 3 Litmus 7 O-cresolphtalein 9.2 Thymolphtalein 10 Indicators change color over a ph range = pka 1 Erythrosine 2 4 Litmus 6 8 o-cresolphthalein 8.2 10.2 Thymolphthalein 9 11 Since this a titration of a weak acid with a strong base the color will change at the ph > 7, (ph at the equivalence point = 8.57). Therefore, o-cresolphthalein is the best indicator because 8.57 lies in the range of color change of o-cresolphthalein.
This is titration curve for 100mL, 0.025M acetic acid with 0.10M NaOH. Choose the best indicator from the following, given their ph range of color change: a. Methyl orange 3.2 4.4 b. Methyl red 4.8 6.0 c. Bromothymol blue 6.1 7.6 d. Phenolphthalein 8.2 10.0 e. Alizarin 11.0 12.4 What part of the curve corresponds to optimum buffer action? Point V 431. The titration curves of polyprotic acid titrations: have as many rises as there are acidic ions. SQ13 SQ13 Sketch the titration curve of H 3 PO 4 with NaOH.
An unknown acid was titrated with a solution of NaOH. The ph titration curve is shown above. Which of the following could the acid be? Acetic acid, CH 3 COOH Hydrochloric acid, HCl Formic acid, HCOOH Nitric acid, HNO 3 Oxalic acid, H 2 C 2 O 4 Boric acid, H 3 BO 3