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Chapter 15 Acid-Base Titration and pH

Chapter 15 Acid-Base Titration and pH. Solution Concentrations*. Molarity – one mole of solute dissolved in enough solvent (water) to make exactly one liter of solution. Molality – one mole of solute dissolved in exactly 1,000 grams of solvent.

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Chapter 15 Acid-Base Titration and pH

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  1. Chapter 15Acid-Base Titration and pH

  2. Solution Concentrations* Molarity – one mole of solute dissolved in enough solvent (water) to make exactly one liter of solution. Molality – one mole of solute dissolved in exactly 1,000 grams of solvent. Normality – one gram equivalent weight (gew) of solute dissolved in enough solvent to make exactly one liter of solution. These are on your Ch 14/15 handout titled: “ph/Acid/Base Equations”

  3. Molarity (M) Formulae M = grams of solute given GFW of solute liters of solvent Grams of solute needed= M(GFW of solute)(Liters Solvent) L of solvent needed = g solute/GFW solute / M

  4. Some molarity problems are on pages 420 and 421 You have 3.50L of solution that contains 90.0g of NaCl. What is the molarity of the solution? g NaCl x 1mol NaCl= mol NaCl, the solute g NaCl mol of Solute = molarity of solution L of solution 90.0g NaCl x 1 mol NaCl= 1.54 mol NaCl 58.44 g NaCl 1.54 mol NaCl / 3.50 L = 0.440 M NaCl

  5. Another molarity problem. P421 practice. • What is the molarity of a solution composed of 5.85g KI, dissolved in enough water to make 0.125 L of solution? 5.85g KI x 1 mol KI = 0.0352 mol KI 166g KI 0.0352 mol KI = 0.282 M KI 0.125 L Complete #2 and # on P421. Check your answers in the back of the book.

  6. molality, m – one mole of solute dissolved in exactly 1,000 g of solvent. m = g of solute given X 1000 GFW of solute X g of solvent g of solute needed = m(GFW)(g solvent) 1000 g solvent needed = g solute(1000) GFW solute x m Molality = molarity if water is the solvent (aqueous solutions)

  7. Normality, N – one gram equivalent weight of solute dissolved in enough solvent to make exactly one liter of solution. N = _________g solute_____ GEW solute x L solvent g solute = N X GEW solute X L of solvent L solvent = ____g of solute___ GEW of solute X N Normality to Molarity M = N(valence of cation)(subscript of cation) GEW = _________GFW of solute_______ charge X subscript of solute cation GEW – gram equivalent weight

  8. pH – What is it? pH is an indication of the hydronium ion concentration present in a solution. [H30+] is the symbol for concentration of hydronium ion in moles per liter or mol/L pOH is an indication of the hydroxide ion concentration present in a solution. [OH-] is the symbol for concentration of hydroxide ion in mol/L

  9. Water self ionizes H20(l) + H20(l) H30+(aq) + OH-(aq) In the above reaction, two water molecules produce a hydronium ion and a hydroxide ion by transfer of a proton. Water is self Ionizing. At 25oC, the concentrations of H30+ and OH- are each only 1.0x10-7 mol/L of water.

  10. Math product of these ions is a constant kw, the ionization constant of water. Kw= [H30+ ] [OH-] = 1.0x10-7(1.0x10-7) =1.0x10-14 This occurs at 25oC. If the temperature changes, the ion product, Kw changes. When both[H30+ ] and[OH-] are 1.0x10-7, the solution is neutral. If [H30+ ] is greater than 1.0x10-7, the solution is Acidic. (10-6 or 10-4 would be greater) If [OH-] is greater than 1.0x10-7, the solution is Basic.

  11. Calculating without a calculator Kw= [H30+ ] [OH-] = 1.0x10-7(1.0x10-7) =1.0x10-14 Let’s say that the [H30+ ] is 1.0x10-6 and you are asked to find the [OH-]. Kw= [H30+ ] [OH-] --> [OH-] = kw = 1.0x10-14 [H30+ ] 1.0x10-6 -14 – (-6) = -14 + 6 = -8 so: [OH-] = 10-8 mol/Liter More practice: 10-14/10-2 = 10-12 and 10-14/10-9 = 10-5

  12. Calculating [H30+ ] and [OH-] Your own scientific calculator is a MUST here!!! Find these keys: 2nd, either EE or EXP, and change sign (-) or (+/-) on your calculator. Let’s practice putting in numbers in sci. not. 1x10-7: Press keys in this sequence: 1 2nd EE (-) 7 on your display you see something similar to this: 1E -7 2 x10-4: 2 2nd EE (-) 4 display: 2 E -4

  13. For concentration, M means moles/L The [H30+] is 2.34 x 10-5 M in a solution. Calculate the [OH-] of the solution. [OH-] = Kw = 1.0x10-14 [H30+ ] 2.34 x 10-5 Key sequence: 1 2nd EE (-) 14 : 2.34 2nd EE (-) 5 enter Display: 4.27 E -10 which means: 4.27 x 10-10 M

  14. Calculate hydronium and hydroxide ion concentrations in a solution that is 1x10-4 M HCl. HCl is a strong acid that ionizes completely. So the concentration of H30+ is 1x10-4 M. Find [OH-]: [OH-] = Kw = 1.0x10-14 [H30+ ] 1x10-4 Answer: [OH-] = 10-10 M Asgn: Page 502 in book: Practice 2,3,4

  15. The pH Scale is used to show how acidic or basic (alkaline) a solution is. pH of a solution is the negative of the common logarithm of the hydronium ion concentration. pH = - log [H30+ ] A common logarithm of a number is “the power to which 10 must be raised to equal the number.”

  16. The logarithm of 1.0x10-7 is - 7.0 The pH = - log [H30+ ] = - log (1.0x10-7) = 7.0 pOH is the negative log of [OH-]. pOH = -log [OH-]. In a neutral solution where [OH-] is 1.0x10-7, the pOH = -log [OH-] = -log 1.0x10-7 = 7.0

  17. Kw= [H30+ ] [OH-] = 1.0x10-7(1.0x10-7) =1.0x10-14 From above: pH + pOH = 14.0 pOH can also be found by: pOH = 14.0 – pH = 14.0 – 7.0 = 7.0

  18. Determine the pH of these solutions: • 1x10-3 M HCl. HCl is a strong acid that ionizes completely. So the concentration of H30+ is 1x10-3 M. pH = - log [H30+ ] = - log (1.0x10-3) = 3.0 • 1x10-4 M NaOH. NaOH is a strong base that ionizes completely. The concentration of OH- is 1x10-4 M [H30+ ] = Kw = 10-14 = 10-10 [OH-] 10-4 pH = -log(10-10) = 10.0

  19. More samples… Find the pH of a solution where [H30+ ] is 2.8 x 10-5 M? pH = -log [H30+ ] = -log 2.8 x 10-5 = 4.55 Key sequence: (-) log 2.8 2nd EE (-) 5 = Find the pH of a 4.7 x 10-2 M NaOH solution. [H30+ ] = Kw = 10-14 = 2.1x10-13 [OH-] 4.7x10-2 pH = -log [H30+ ] = - log 2.1x10-13 = 12.7

  20. Assignment • 505/1. b and d • 506/1-4

  21. Review • Acid + Base --> Salt + Water • The mixture is neutralized (no longer acidic or basic).

  22. In this section we are going to look at: indicators, pH meters, and titrations. You used 2 indictors in a lab recently to determine how acidic or basic several solutions were. You used litmus paper and pH paper (Hydrion) Acid-base indicators are sensitive to pH of acids and bases. They will change color as a result of the ions present.

  23. In acidic solutions, an indicator will be one color (litmus turns red) and in basic solutions, an indicator will be another color (litmus burns blue).

  24. Indicator Samples Methyl red, Bromthymol blue, Methyl orange, Phenolphthalein, Phenol red are indicator samples. These will ionize in solution and, depending upon their acid or base strength will change color over a range of pH values until the end point is reached. The range over which an indicator changes color is called its transition interval.

  25. Reading Indicator Values Litmus gives a very broad reading – a solution is either acidic or basic. Indicators are more specific in reading the pH of an acid or base. But the most accurate method of measuring pH is with a pH meter. A pH meter determines the pH of a solution by electrically measuring the voltage between the two electrodes placed in a solution.

  26. Titration Titration is the controlled addition and measurement of the amount of a solution of known concentration required to react completely with a measured amount of a solution of unknown concentration. More simply: it is using a known concentration of a solution to determine the concentration of a solution of unknown concentration.

  27. When a base is added to an acid the solution will become neutral and this will be shown by an indicator changing solor. This will occur when equal numbers of H30+ and OH- are present. Picture retrieved from: cikguwong.blogspot.com

  28. The figure below shows typical pH curves for various acid-base titrations. The equivalence points and end points are different for the various combinations of strong and weak acids and bases. Equivalence point The point at which two solutions used in a titration are present in chemically equivalent amounts. End point The point in a titration at which an indicator changes color. Retrieved from: chemguide.co.uk

  29. Determination of the acidity/alkalinity of salt solutions produced by neutralization reactions. • The relative pH of a neutralized solution can be determined utilizing the following scale showing A/B strengths. pH 1-3 strong acid with a weak base pH 3-5 strong acid with a moderate base pH 5-7 moderate acid with a weak base pH 7(neutral)equal strength acid/base reacton pH 7-9 moderate base with a weak acid pH 9-12 strong base with a moderate acid pH 12-14 strong base with a weak acid See your Ch 14/15 handout for more information.

  30. Molarity and Titration • Standard solution – the solution that contains the precisely known concentration of a solute. • Primary standard – highly purified solid compound used to check the concentration of the known solution in a titration. • Knowing the molarity and volume of a known solution used in a titration, the molarity of a given volume of a solution with unknown concentration can be found.

  31. Titration Set up and Procedure Your book has a more complete explanation. Fill one buret with an acid. Record volume. Fill other buret with standard solution base. Record volume. Indicator (Phenolphthalein) will be in a flask. Add a given amount of A to the flask. Begin adding B to the flask until the pink color of the indicator begins to form. Swirl the contents constantly. As the pink color begins to remain for longer periods of time, you are nearing the end point. When the pink color remains after 30 seconds of swirling, the equivalence point is reached. Record the exact volume of the base put in the flask. Retrieved from web.ysu.edu.

  32. Formula to calculate molarity in a titration. Ma x Va = Mb x Vb Ma – molarity of acid Va – volume of acid Mb – molarity of base Vb – volume of base See Ch 14/15 handout for more info.

  33. Molarity & Titation Problem Steps 1. Start with balanced equation for the neutralization reaction, and determine the chemically equivalent amount of the acid and base.

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