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

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Presentation on theme: "Chapter FifteenPrentice-Hall ©2002Slide 1 of 31 920131 1 48slides."— Presentation transcript:

1 Chapter FifteenPrentice-Hall ©2002Slide 1 of slides

2 Chapter FifteenPrentice-Hall ©2002Slide 2 of 31 Neutralization Reactions slides

3 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) slides

4 Chapter FifteenPrentice-Hall ©2002Slide 4 of slides

5 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 indicatoracids basespKa slides

6 Chapter FifteenPrentice-Hall ©2002Slide 6 of slides

7 Chapter FifteenPrentice-Hall ©2002Slide 7 of slides

8 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 slides

9 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 slides

10 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 slides

11 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  slides

12 Chapter FifteenPrentice-Hall ©2002Slide 12 of slides

13 Chapter FifteenPrentice-Hall ©2002Slide 13 of slides

14 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 slides

15 Chapter FifteenPrentice-Hall ©2002Slide 15 of slides

16 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 http"\\asadipour.kmu.ac.ir slides 16

17 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 http"\\asadipour.kmu.ac.ir slides 17

18 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? http"\\asadipour.kmu.ac.ir slides 18

19 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 mL of M HCl with 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 + ] http"\\asadipour.kmu.ac.ir slides 19

20 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 mL of M HCl with M NaOH. b)equivalence point. Answer Q1. There are:NaCl & H 2 O Answer Q2. H 2 O Answer Q3. Answer Q4. pH= http"\\asadipour.kmu.ac.ir slides 20

21 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 mL of M HCl with 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 + ] http"\\asadipour.kmu.ac.ir slides 21

22 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 mL of M HCl with 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 http"\\asadipour.kmu.ac.ir slides 22

23 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 http"\\asadipour.kmu.ac.ir slides 23

24 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× essential points. 1)initial point 2)equivalence point 3)beyond the initial point 4)before the equivalence point 5)beyond the equivalence point http"\\asadipour.kmu.ac.ir slides 24

25 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? http"\\asadipour.kmu.ac.ir slides 25

26 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 mL of M CH 3 COOH with 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 + ] http"\\asadipour.kmu.ac.ir slides 26

27 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 mL of M CH 3 COOH with 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 http"\\asadipour.kmu.ac.ir slides 27

28 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 mL of M CH 3 COOH with 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 Q http"\\asadipour.kmu.ac.ir slides 28

29 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 mL of M CH 3 COOH with 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 Q http"\\asadipour.kmu.ac.ir slides 29

30 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 mL of M CH 3 COOH with 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 http"\\asadipour.kmu.ac.ir slides 30

31 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 http"\\asadipour.kmu.ac.ir slides 31

32 Chapter FifteenPrentice-Hall ©2002Slide 32 of slides

33 Chapter FifteenPrentice-Hall ©2002Slide 33 of 31 Titration curves for HCl with NaOH. A ml of M HCl with M NaOH. B ml of M HCl with M NaOH slides

34 Chapter FifteenPrentice-Hall ©2002Slide 34 of 31 Titration curves for the titration of acetic acid with NaOH. A M acetic acid with M NaOH. B M acetic acid with M NaOH slides

35 Chapter FifteenPrentice-Hall ©2002Slide 35 of 31 Titration curves for the titration of acetic acid with NaOH. A M acetic acid with M NaOH. B M acetic acid with M NaOH. Titration curves for HCl with NaOH. A ml of M HCl with M NaOH. B ml of M HCl with M NaOH slides

36 Chapter FifteenPrentice-Hall ©2002Slide 36 of 31 General Shapes of Titration Curves Effect of pKa Effect of initial concentration slides

37 Chapter FifteenPrentice-Hall ©2002Slide 37 of 31 The effect of acid strength (dissociation constant) on titration curves. Each curve represents the titration of ml of M acid with M base. Effect of Ka slides

38 Chapter FifteenPrentice-Hall ©2002Slide 38 of 31 The effect of base strength (dissociation constant) on titration curves. Each curve represents the titration of ml of M base with M HCl. Effect of Kb slides

39 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 slides

40 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 NH 3 + H 2 O  NH OH - K 2 = K b = 1.8 x H 3 O + + OH -  2 H 2 O K 3 = 1/ K w = 1 x 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 slides

41 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 ? 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 neutral- ized 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 neutral- ize a solution containing 32.0 g NaOH? slides

42 Chapter FifteenPrentice-Hall ©2002Slide 42 of 31 Titration problems 1. (3)(0.15 M)( L) = (1)(0.10 M)(V B ) V B = (3)(0.15 M)( L) / (1)(0.10 M) = 0.11 L 2. (1)(M A )( L) = (2)(0.10 M)(0.040 L) M A = (2)(0.10 M)(0.040 L) / (1)( 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 = (2)(1.50 mol/L)(V A ) = (1)(0.800 mol) V A = (1)(0.800 mol) / (2)(1.50 mol/L) = L slides

43 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 slides

44 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 = slides

45 Chapter FifteenPrentice-Hall ©2002Slide 45 of 31 Plots of relative amounts of acetic acid and acetate ion during a titration slides


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