Chapter FifteenPrentice-Hall ©2002Slide 1 of 31 920131 1 48slides.

Chapter FifteenPrentice-Hall ©2002Slide 3 of 31 Standard Solutions: strong acids or strong bases because they will react completely – Acids: (HCl), (HClO 4 ), (H 2 SO 4 ) – Bases: (NaOH), (KOH) 920131 3 http:\\asadipour.kmu.ac.ir 48slides

Chapter FifteenPrentice-Hall ©2002Slide 5 of 31 Thymol blue (thymolsulphonepht halein) is a brownish-green or reddish-brown crystaline powder that is used as an pH indicator. It is insoluble in water but soluble in alcohol and dilute alkali solutions. It transitions from red to yellow at pH 1.2– 2.8 and from yellow to blue from at pH 8.0–9.6.pH indicatorwater alcohol alkalipH Bromophenol blue (3',3",5',5"- tetrabromophenol sulfonphthalein) is an acid-base indicator whose useful range as an indicator lies between pH 3.0 and 4.6. It changes from yellow at pH 3.0 to purple at pH 4.6; this reaction is reversible. indicatorpH Chloropheno l red is an indicator dye that changes color from yellow to violet in the pH range 4.8 to 6.7. The lamda max is at 572 nm. lamda max A solution of phenol red is used as a pH indicator: its color exhibits a gradual transition from yellow to red over the pH range 6.6 to 8.0. Above pH 8.1, phenol red turns a bright pink (fuchsia) color. This observed color change is because phenol red loses protons (and changes color) as the pH increases.pH indicatorfuchsia Bromothymol Blue (also known as dibromothymol sulfonephthalei n, Bromthymol Blue, and BTB) is a chemical indicator for weak acids and bases The pKa for bromothymol blue is 7.10. indicatoracids basespKa 920131 5 http:\\asadipour.kmu.ac.ir 48slides

Chapter FifteenPrentice-Hall ©2002Slide 8 of 31 Acid/Base Indicators Many substances display colors that depend on the pH of the solutions in which they are dissolved. An acid/base indicator is a weak organic acid or a weak organic base whose undissociated form differs in color from its conjugate form. e.g., the behavior of an acid-type indicator, HIn, is described by the equilibrium HIn + H 2 O In - + H 3 O + acid color base color The equilibrium for a base-type indicator, In, is In + H 2 O InH + + OH - base coloracid color 920131 8 http:\\asadipour.kmu.ac.ir 48slides

Chapter FifteenPrentice-Hall ©2002Slide 9 of 31 …continued… The equilibrium-constant expression for the dissociation of an acid-type indicator takes the form Rearranging leads to The hydronium ion concentration determines the ratio of the acid to the conjugate base form of the indicator and thus determines the color developed by the solution. + - + - 920131 9 http:\\asadipour.kmu.ac.ir 48slides

Chapter FifteenPrentice-Hall ©2002Slide 10 of 31 …continued… The color imparted to a solution by a typical indicator appears to the average observer to change rapidly only within the limited concentration ratio of approximately 10 to 0.1 and its base color when The color appears to be intermediate for ratios between these two values. These ratios vary considerably from indicator to indicator. 920131 10 http:\\asadipour.kmu.ac.ir 48slides

Chapter FifteenPrentice-Hall ©2002Slide 11 of 31 …continued… For the full acid color, [H 3 O + ] = 10K a and similarly for the full base color, [H 3 O + ] = 0.1K a To obtain the indicator pH range, we take the negative logarithms of the two expression: pH (acid color) = -log (10K a ) = pK a - 1 pH (basic color) = -log (0.1K a ) = pK a + 1 indicator pH range = pK a  1 920131 11 http:\\asadipour.kmu.ac.ir 48slides

Chapter FifteenPrentice-Hall ©2002Slide 14 of 31 Variables: 1)temperature,2)ionic strength of medium 3)presence of organic solvents 4)presence of colloidal particles 920131 14 http:\\asadipour.kmu.ac.ir 48slides

Chapter FifteenPrentice-Hall ©2002Slide 16 of 31 Neutralization Reactions Neutralization is the reaction of an acid and a base. Titration is a common technique for conducting a neutralization. At the equivalence point in a titration, the acid and base have been brought together in exact stoichiometric proportions. The point in the titration at which the indicator changes color is called the end point. The indicator endpoint and the equivalence point for a neutralization reaction can be best matched by plotting a titration curve, a graph of pH versus volume of titrant. In a typical titration, 50 mL or less of titrant that is 1 M or less is used. 920118 http"\\asadipour.kmu.ac.ir........52 slides 16

Chapter FifteenPrentice-Hall ©2002Slide 17 of 31 Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O Calculate the pH at the some points and draw the curve. 4 essential points. 1)initial point 2)equivalence point 3)before the equivalence point 4)beyond the equivalence point Ml تیترانت pH محیط 0 15 19 19.5 19.9 20 20.1 20.5 21 25 40 920118 http"\\asadipour.kmu.ac.ir........52 slides 17

Chapter FifteenPrentice-Hall ©2002Slide 18 of 31 Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O 4 questions. 1)What are the present compounds? 2)Which of them is effective on pH? 3)How much are the concentrations? 4)What is the relationship between their Conc. And pH? 920118 http"\\asadipour.kmu.ac.ir........52 slides 18

Chapter FifteenPrentice-Hall ©2002Slide 19 of 31 Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH. (a)initial pH. (Before the addition of any NaOH). Answer Q1. There are:HCl & H2O Answer Q2. HCl Answer Q3. [HCl] Answer Q4. pH=-log[H + ] 920118 http"\\asadipour.kmu.ac.ir........52 slides 19

Chapter FifteenPrentice-Hall ©2002Slide 20 of 31 Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH. b)equivalence point. Answer Q1. There are:NaCl & H 2 O Answer Q2. H 2 O Answer Q3. Answer Q4. pH=7 920118 http"\\asadipour.kmu.ac.ir........52 slides 20

Chapter FifteenPrentice-Hall ©2002Slide 21 of 31 Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH. c)before the equivalence point. Answer Q1. There are:HCl,NaCl & H 2 O Answer Q2. HCl Answer Q3. Answer Q4. [H + ]=N pH=-log[H + ] 920118 http"\\asadipour.kmu.ac.ir........52 slides 21

Chapter FifteenPrentice-Hall ©2002Slide 22 of 31 Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH. d)after the equivalence point. Answer Q1. There are:NaOH,NaCl & H 2 O Answer Q2. NaOH Answer Q3. Answer Q4. [OH - ]=N pOH=-log[OH - ] pH=14-pOH 920118 http"\\asadipour.kmu.ac.ir........52 slides 22

Chapter FifteenPrentice-Hall ©2002Slide 23 of 31 Titration Curve For Strong Acid - Strong Base pH is low at the beginning. pH changes slowly until just before equivalence point. pH changes sharply around equivalence point. pH = 7.0 at equivalence point. Further beyond equivalence point, pH changes slowly. Any indicator whose color changes in pH range of 4 – 10 can be used in titration. 920118 http"\\asadipour.kmu.ac.ir........52 slides 23

Chapter FifteenPrentice-Hall ©2002Slide 24 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the some points and draw the curve. K a =1×10 -5 5 essential points. 1)initial point 2)equivalence point 3)beyond the initial point 4)before the equivalence point 5)beyond the equivalence point 920118 http"\\asadipour.kmu.ac.ir........52 slides 24

Chapter FifteenPrentice-Hall ©2002Slide 25 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O 4 questions. 1)What are the present compounds? 2)Which of them is effective on pH? 3)How much are the concentrations? 4)What is the relationship between their Conc. And pH? 920118 http"\\asadipour.kmu.ac.ir........52 slides 25

Chapter FifteenPrentice-Hall ©2002Slide 26 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH with 0.500 M NaOH. (a)initial pH. (Before the addition of any NaOH). Answer Q1. There are: CH 3 COOH & H 2 O Answer Q2. CH 3 OOH Answer Q3. CH 3 OOH Answer Q4. pH=-log[H + ] 920118 http"\\asadipour.kmu.ac.ir........52 slides 26

Chapter FifteenPrentice-Hall ©2002Slide 27 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH with 0.500 M NaOH. b)equivalence point. Answer Q1. There are: CH3COO -, Na + & H 2 O Answer Q2. CH3COO - Answer Q3. Answer Q4. pOH=-log[OH - ] K a ×K b =K w 920118 http"\\asadipour.kmu.ac.ir........52 slides 27

Chapter FifteenPrentice-Hall ©2002Slide 28 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH with 0.500 M NaOH. c)beyond the initial point. Answer Q1. There are: CH 3 COOH, CH3COO -,Na + & H 2 O Answer Q2. CH 3 COOH, CH3COO - Answer Q3. Answer Q4. 920118 http"\\asadipour.kmu.ac.ir........52 slides 28

Chapter FifteenPrentice-Hall ©2002Slide 29 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH with 0.500 M NaOH. d)before the equivalence point. Answer Q1. There are: CH 3 COOH, CH3COO -,Na + & H 2 O Answer Q2. CH 3 COOH, CH3COO - Answer Q3. Answer Q4. 920118 http"\\asadipour.kmu.ac.ir........52 slides 29

Chapter FifteenPrentice-Hall ©2002Slide 30 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH with 0.500 M NaOH. e)after the equivalence point. Answer Q1. There are:NaOH, CH3COO -, Na + & H 2 O Answer Q2. NaOH Answer Q3. Answer Q4. [OH - ]=N pOH=-log[OH - ] pH=14-pOH 920118 http"\\asadipour.kmu.ac.ir........52 slides 30

Chapter FifteenPrentice-Hall ©2002Slide 31 of 31 Titration Curve For Weak Acid - Strong Base The initial pH is higher because weak acid is partially ionized. At the half-neutralization point, pH = pKa. pH >7 at equivalence point because the anion of the weak acid hydrolyzes. The steep portion of titration curve around equivalence point has a smaller pH range. The choice of indicators for the titration is more limited. 920118 http"\\asadipour.kmu.ac.ir........52 slides 31

Chapter FifteenPrentice-Hall ©2002Slide 33 of 31 Titration curves for HCl with NaOH. A. 50.0 ml of 0.0500 M HCl with 0.1000M NaOH. B. 50.00 ml of 0.000500 M HCl with 0.00100 M NaOH. 920131 33 http:\\asadipour.kmu.ac.ir 48slides

Chapter FifteenPrentice-Hall ©2002Slide 34 of 31 Titration curves for the titration of acetic acid with NaOH. A. 0.1000 M acetic acid with 0.1000M NaOH. B. 0.001000 M acetic acid with 0.00100 M NaOH. 920131 34 http:\\asadipour.kmu.ac.ir 48slides

Chapter FifteenPrentice-Hall ©2002Slide 35 of 31 Titration curves for the titration of acetic acid with NaOH. A. 0.1000 M acetic acid with 0.1000M NaOH. B. 0.001000 M acetic acid with 0.00100 M NaOH. Titration curves for HCl with NaOH. A. 50.0 ml of 0.0500 M HCl with 0.1000M NaOH. B. 50.00 ml of 0.000500 M HCl with 0.00100 M NaOH. 920131 35 http:\\asadipour.kmu.ac.ir 48slides

Chapter FifteenPrentice-Hall ©2002Slide 37 of 31 The effect of acid strength (dissociation constant) on titration curves. Each curve represents the titration of 50.00 ml of 0.1000 M acid with 0.1000 M base. Effect of Ka 920131 37 http:\\asadipour.kmu.ac.ir 48slides

Chapter FifteenPrentice-Hall ©2002Slide 38 of 31 The effect of base strength (dissociation constant) on titration curves. Each curve represents the titration of 50.00 ml of 0.1000 M base with 0.1000 M HCl. Effect of Kb 920131 38 http:\\asadipour.kmu.ac.ir 48slides

Chapter FifteenPrentice-Hall ©2002Slide 39 of 31 non-aqueous acid base titration 1)Solubility 2) acid or base strength Acid and base strengths that are not distinguished in aqueous solution may be distinguishable in non-aqueous solvents. HClO 4 > HCl in acetic acid solvent, neither acid is completely dissociated. HClO 4 + CH 3 COOH  ClO 4 – + CH 3 COOH 2 + K = 1.3×10 –5 strong acid strong base weak base weak acid HCl + CH 3 COOH  Cl – + CH 3 COOH 2 + K = 5.8×10 –8 Differentiate acidity or basicity of different acids or bases differentiating solvent for acids …… acetic acid differentiating sovent for bases …… ammonia 920131 39 http:\\asadipour.kmu.ac.ir 48slides

Chapter FifteenPrentice-Hall ©2002Slide 40 of 31 Reaction between weak acid and weak base –Both the weak acid and weak base remain largely undissociated and neutralization involves proton transfer from the weak acid to the weak base. Consider acetic acid and ammonia: CH 3 COOH + NH 3  CH 3 COO - + NH 4 + is composed of CH 3 COOH + H 2 O  CH 3 COO - + H 3 O + K 1 = K a = 1.8 x 10 -5 NH 3 + H 2 O  NH 4 + + OH - K 2 = K b = 1.8 x 10 -5 H 3 O + + OH -  2 H 2 O K 3 = 1/ K w = 1 x 10 14 K n = K overall = K 1 x K 2 x K 3 = K b K a / K w = 3.2 x 10 4 Therefore, the Reaction Still Shifts significantly to the right -- ionizing much of the component present in the smaller amount 920131 40 http:\\asadipour.kmu.ac.ir 48slides

Chapter FifteenPrentice-Hall ©2002Slide 41 of 31 Titration problems 1.What volume of 0.10 mol/L NaOH is needed to neutralize 25.0 mL of 0.15 mol/L H 3 PO 4 ? 2.25.0 mL of HCl(aq) was neutralized by 40.0 mL of 0.10 mol/L Ca(OH) 2 solution. What was the concentration of HCl? 3.A truck carrying sulfuric acid is in an accident. A laboratory analyzes a sample of the spilled acid and finds that 20 mL of acid is neutralized by 60 mL of 4.0 mol/L NaOH solution. What is the concentration of the acid? 4.What volume of 1.50 mol/L H 2 S will neutralize a solution containing 32.0 g NaOH? 920131 41 http:\\asadipour.kmu.ac.ir 48slides

Chapter FifteenPrentice-Hall ©2002Slide 42 of 31 Titration problems 1. (3)(0.15 M)(0.0250 L) = (1)(0.10 M)(V B ) V B = (3)(0.15 M)(0.0250 L) / (1)(0.10 M) = 0.11 L 2. (1)(M A )(0.0250 L) = (2)(0.10 M)(0.040 L) M A = (2)(0.10 M)(0.040 L) / (1)(0.0250 L) = 0.32 M 3. Sulfuric acid = H 2 SO 4 (2)(M A )(0.020 L) = (1)(4.0 mol/L)(0.060 L) M A = (1)(4.0 M)(0.060 L) / (2)(0.020 L) = 6.0 M 4. mol NaOH = 32.0 g x 1 mol/40.00 g = 0.800 (2)(1.50 mol/L)(V A ) = (1)(0.800 mol) V A = (1)(0.800 mol) / (2)(1.50 mol/L) = 0.267 L 920131 42 http:\\asadipour.kmu.ac.ir 48slides

Chapter FifteenPrentice-Hall ©2002Slide 43 of 31 Species concentrations of weak diprotic acids Evaluate concentrations of species in a 0.10 M H 2 S solution. Solution: H 2 S = H + + HS – K a1 = 1.02e-7 (0.10–x) x+y x-yAssume x = [HS – ] HS – = H + + S 2– K a2 = 1.0e-13 x–y x+y yAssume y = [S 2– ] (x+y) (x-y) (x+y) y ————— = 1.02e-7 ———— = 1.0e-13 (0.10-x)(x-y) [H 2 S] = 0.10 – x = 0.10 M [HS – ] = [H + ] = x  y = 1.0e–4 M; [S 2– ] = y = 1.0e-13 M 0.1>> x >> y: x+ y = x-y = x x =  0.1*1.02e-7 = 1.00e-4 y = 1e-13 920131 43 http:\\asadipour.kmu.ac.ir 48slides

Chapter FifteenPrentice-Hall ©2002Slide 44 of 31 Alpha Values Def.: the relative equilibrium concentration of the weak acid/base and its conjugate base/acid (titrating HOAc with NaOH): α 0 + α 1 = 1 920131 44 http:\\asadipour.kmu.ac.ir 48slides

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