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CONCEPT: CLASSIFICATION AND IDENTIFICATION OF BUFFERS Solutions which contain a acid and its base are called buffer solutions because they resist drastic changes in ph. They resist drastic changes in ph by keeping and constant. Adding a small amount of STRONG BASE and, the ph, but not by much because the neutralizes the STRONG BASE added. Adding a small amount of STRONG ACID and, the ph, but not by much because the neutralizes the STRONG ACID added. PRACTICE 1: Which one of the following combinations does not create a buffer? a) HC2H3O2 and K C2H3O2 b) H2SO3 and NaHSO3 c) H3PO4 and NaH2PO4 d) HNO3 and KNO3 e) NH4Cl and NH3 PRACTICE 2: Which of the following combinations can result in the formation of a buffer? a) HF and HI b) HC2H3O2 and NH3 c) CH3CH2NH2 and CH3CH2NH3 + d) NaCl and NaOH Page 2
CONCEPT: CREATING A BUFFER There are 3 ways to form a buffer: 1) Mixing a acid and its base. 2) Mixing a acid and a base. 3) Mixing a acid and a base. Page 3
PRACTICE: CREATING A BUFFER EXAMPLE: Which of the following combinations can result in the formation of a buffer? a) 0.01 moles HClO (hypochlorous acid) and 0.05 moles of NaOH. b) 0.01 moles HClO (hypochlorous acid) and 0.05 moles of HCl. c) 0.01 moles HClO (hypochlorous acid) and 0.05 moles of NH3. d) 0.01 moles HClO (hypochlorous acid) and 0.001 moles of NaOH PRACTICE 1: Which of the following combinations can result in the formation of a buffer? a) 50 ml of 0.10 M HF with 50 ml of 0.10 M NaOH. b) 50 ml of 0.10 M HNO2 with 25 ml of 0.10 M Ca(OH)2. c) 50 ml of 0.10 M CH3CO2H with 60 ml of 0.10 M NaOH. d) 50 ml of 0.10 M HF with 30 ml of 0.10 M NaOH. PRACTICE 2: A buffer solution is comprised of 50.0 ml of a 0.100 M HC2H3O2 and 60.0 ml of a 0.100 M NaC2H3O2. Which of the following actions would completely destroy the buffer? a) Adding 0.003 mol HC2H3O2 b) Adding 0.007 mol Ca(C2H3O2)2 c) Adding 0.005 mol NaOH d) Adding 0.004 mol HCl e) Adding 0.001 mol HCl Page 4
CONCEPT: CALCULATING THE ph OF BUFFERS We learned that whenever we had a(n) acid or base we were supposed to use our favorite friend the Chart in order to calculate the ph or poh. Now, whenever we have a buffer solution we can skip it and use the Equation. Buffer Equation: (conjugate base) ph = pka + log (weak acid) EXAMPLE 1: What is the ph of a solution consisting of 2.75 M sodium phenolate (C6H5ONa) and 3.0 M phenol (C6H5OH). The Ka of phenol is 1.0 x 10-10. PRACTICE: Calculate the ph of a solution formed by mixing 200 ml of a 0.400 M C2H5NH2 solution with 350 ml of a 0.450 M C2H5NH3 + solution. (Kb of C2H5NH2 is 5.6 x 10-4 ). Page 5
PRACTICE: CALCULATING THE ph OF BUFFERS (PART 1) EXAMPLE 1: What is the buffer component concentration ratio, Pr, of a buffer that has a ph of 5.11. (The Ka of HPr is H Pr 1.30 x 10-5 ). EXAMPLE 2: Over what ph range will an oxalic acid (H2C2O4) / sodium oxalate (NaHC2O4) solution work most effectively? The acid dissociation constant of oxalic acid is 6.0 x 10-2. a) 0.22 2.22 b) 1.00 3.00 c) 0.22 1.22 d) 2.0 4.0 PRACTICE: Determine how many grams of sodium acetate, NaCH3CO2 (MW: 82.05 g/mol), you would mix into enough 0.065 M acetic acid CH3CO2H (MW: 60.05 g/mol) to prepare 3.2 L of a buffer with a ph of 4.58. The Ka is 1.8 x 10-5. Page 6
PRACTICE: CALCULATING THE ph OF BUFFERS (PART 2) EXAMPLE: Which weak acid-conjugate base combination would be ideal to form a buffer with a ph of 4.74. a) Cyanic acid and Potassium cynate (Ka = 4.9 x 10-10 ) b) Benzoic acid and Lithium benzoate (Ka = 6.3 x 10-5 ) c) Acetic acid and Sodium acetate (Ka = 1.7 x 10-5 ) d) Ammonium chloride and Ammonia (Ka = 5.56 x 10-10 ) e) Formic acid and Cesium formate (Ka = 1.7 x 10-4 ) PRACTICE: A buffer solution is made by combining a weak acid with its conjugate salt. What will happen to the ph if the solution is diluted to one-fourth of its original concentration? a) The ph will increase. b) The ph will decrease. c) The ph will remain constant. d) The solution will become more neutral. Page 7
CONCEPT: ph TITRATION CURVES The shape of a ph titration curve makes it possible to identify the equivalence point, the point at which of acid and base are mixed together. At the equivalence point, the ph of a strong acid and strong base is 7. At the equivalence point, the ph of a strong acid and weak base is 7. At the equivalence point, the ph of a weak acid and strong base is 7. Page 8
PRACTICE: ph TITRATION CURVES (CALCULATIONS 1) EXAMPLE: The following questions refer to the titration curve given below. a) The titration curve shows the titration of a strong acid a weak acid a strong base a weak base with a strong base with a strong base with a strong acid with a strong acid b) Which point on the titration curve represents a region where a buffer solution has formed? point A point B point C point D c) Which point on the titration curve represents the equivalence point? point A point B point C point D d) Which of the following would be the best indicator to use in the titration? erythrosin B methyl red bromthymol blue o-cresonphthalein pka = 2.9 pka = 5.4 pka = 6.8 pka = 9.0 Page 9
PRACTICE: ph TITRATION CURVES (CALCULATIONS 2) EXAMPLE 1: The acid form of an indicator is red and its anion is blue. The Ka value for this indicator is 10-9. What will be the approximate ph range over which this indicator changes color? a) 3-5 b) 4-6 c) 5-7 d) 8-10 e) 9-11 PRACTICE : What will be the color of the indicator in the above question in a solution that has a ph of 6? EXAMPLE 2: Consider the titration of 100.0 ml of 0.016 M H2SO4 with 0.400 M NaOH at the equivalence point. How many many milliters of 0.400 M NaOH are required to reach the equivalence point? Page 10
PRACTICE: ph TITRATION CURVES (CALCULATIONS 3) EXAMPLE: Consider the titration of 40.0 ml of 0.0800 M HCl with 0.0160 M Ca(NH2)2. a) How many milliliters of 0.0160 M Ca(NH2)2 are required to reach the equivalence point? b) What is the ph of this solution? PRACTICE: Consider the titration of 60.0 ml of 0.200 M a H3PO3 solution with 0.350 M potassium hydroxide, KOH solution. How many milliliters of base would be required to reach each of its equivalence points? Page 11
CONCEPT: WEAK ACID & STRONG BASE TITRATIONS In the past, we reacted WEAK Acids or WEAK Bases with and used a(n) Chart. Remember in this case the units in this chart are in. Now we will react WEAK Acids with bases. When you react a WEAK species with a species you have to use a(n) Chart. In this case the units must be in. HNO 2 NaOH Weak Acid Strong Base Before Equivalence Point Initial Change Final HNO 2 NaOH NaNO 2 Weak Acid Strong Base Conjugate Base 0.250 moles H 2 O and will be present at the end. Use the Henderson Hasselbalch Equation. Conjugate Base ph = pka + log Weak Acid After Equivalence Point HNO 2 NaOH NaNO 2 Weak Acid Strong Base Conjugate Base Initial 0.250 moles Change H 2 O will be present at the end. moles left [SB] = poh = log[sb] ph =14 poh Total Liters Final At Equivalence Point HNO 2 NaOH NaNO 2 Weak Acid Strong Base Conjugate Base Initial 0.250 moles Change H 2 O Only will be present at the end. Use an ICE Chart to find ph. [CB] = moles left Total Liters K b = x2 [CB] Final Page 12
PRACTICE: WEAK ACID & STRONG BASE TITRATIONS CALCULATIONS 1 EXAMPLE: Consider the titration of 75.0 ml of 0.0300 M H3C3O3 (Ka = 4.1 X 10-3 ) with 12.0 ml of 0.0450 M KOH. Calculate the ph. PRACTICE: In order to create a buffer, 7.510 g of sodium cyanide is mixed with 100.0 ml of 0.250 M hydrocyanic acid, HCN. What is the ph of the buffer solution after the addition of 175.0 ml of 0.300 M NaH? Page 13
PRACTICE: WEAK ACID & STRONG BASE TITRATIONS CALCULATIONS 2 EXAMPLE: Consider the titration of 75.0 ml of 0.60 M HNO2 with 0.100 M NaOH at the equivalence point. What would be the ph of the solution at the equivalence point? The Ka of HNO2 is 4.6 x 10-4. Page 14
CONCEPT: WEAK BASE & STRONG ACID TITRATIONS Now we will react WEAK Bases with STRONG Acids. Remember when you react a WEAK species with a species you have to use a(n) Chart. Again, in this case the units must be in. NH 3 HBr Before Equivalence Point NH 3 HBr NH + 4 Br Conjugate Base Strong Acid Weak Acid Initial 0.250 moles Change Final and will be present at the end. Use the Henderson Hasselbalch Equation. Conjugate Base ph = pka + log Weak Acid After Equivalence Point NH 3 HBr NH + 4 Br Conjugate Base Strong Acid Weak Acid Initial 0.250 moles Change will be present at the end. [SA] = moles left Total Liters ph = log[sa] Final At Equivalence Point NH 3 HBr NH + 4 Br Conjugate Base Strong Acid Weak Acid Only will be present at the end. Use an ICE Chart to find ph. Initial Change 0.250 moles [WA] = moles left Total Liters K a = x2 [CB] Final Page 15
PRACTICE: WEAK BASE & STRONG ACID TITRATIONS CALCULATIONS EXAMPLE: A buffer contains 167.2 ml of 0.25 M propanoic acid, CH3CH2COOH, with 138.7 ml of 0.42 M sodium propanoate, CH3CH2COONa. Find the ph after the addition of 150.2 ml of 0.56 M HCl. The Ka of CH3CH2COOH is 1.3 x 10-5. PRACTICE: Calculate the ph of the solution that results from the mixing of 75.0 ml of 0.100 M NaC2H3O2 and 75.0 ml of 0.150 M HC2H3O2 with 0.0025 moles of HBr. Ka of HC2H3O2 is 1.8 x 10-5. Page 16
CONCEPT: STRONG ACID & BASE TITRATIONS Whenever we had a STRONG ACID or STRONG BASE we use an ICE CHART. Now, whenever you titrate a STRONG ACID with STRONG BASE you use an ICF CHART. EXAMPLE: Calculate the ph of the solution resulting from the titration of 75.0 ml of 0.100 M HBrO4 with 55.0 ml of 0.100 M NaNH2. PRACTICE: Calculate the ph of the solution resulting from the mixing of 175.0 ml of 0.250 M HNO3 with 75.0 ml of 0.200 M Ba(OH)2. Page 17
CONCEPT: DIPROTIC & POLYPROTIC BUFFERS A diprotic or polyprotic buffer can be approached in a way similar to monoprotic buffers. The key difference is that multiple pka values will be involved. For Monoprotic Buffers HClO + NaClO (aq) Weak oxyacid Conjugate base (conjugate base) ph = pka + log (weak acid) (Has 1 less Hydrogen) For Diprotic Buffers H 2 A HA A 2 ph = ph = For Polyprotic Buffers H 3 A H 2 A HA 2 A 3 ph = ph = ph = Page 18
PRACTICE: DIPROTIC & POLYPROTIC BUFFERS CALCULATIONS 1 EXAMPLE 1: Calculate the ph of 100 ml of a 0.25 M H2CO3 when 70.0 ml of 0.25 M NaOH are added. Ka1 = 4.3 x 10-7 and Ka2 = 5.6 x 10-11. EXAMPLE 2: Calculate the ph of 75.0 ml of a 0.10 M of phosphorous acid, H3PO3, when 80.0 ml of 0.15 M NaOH are added. Ka1 = 5.0 x 10-2, Ka2 = 2.0 x 10-7. Page 19