Presentation is loading. Please wait.

Presentation is loading. Please wait.

Hydronium Ions and Hydroxide Ions Self-Ionization of Water In the self-ionization of water, two water molecules produce a hydronium ion and a hydroxide.

Similar presentations


Presentation on theme: "Hydronium Ions and Hydroxide Ions Self-Ionization of Water In the self-ionization of water, two water molecules produce a hydronium ion and a hydroxide."— Presentation transcript:

1 Hydronium Ions and Hydroxide Ions Self-Ionization of Water In the self-ionization of water, two water molecules produce a hydronium ion and a hydroxide ion by transfer of a proton. Chapter 15 Section 1 Aqueous Solutions and the Concept of pH In water at 25°C, [H 3 O + ] = 1.0 ×10 −7 M and [OH − ] = 1.0 × 10 −7 M. The ionization constant of water, K w, is expressed by the following equation. K w = [H 3 O + ][OH − ]

2 Hydronium Ions and Hydroxide Ions, continued Neutral, Acidic, and Basic Solutions Solutions in which [H 3 O + ] = [OH − ] is neutral. Solutions in which the [H 3 O + ] > [OH − ] are acidic. [H 3 O + ] > 1.0 × 10 −7 M Solutions in which the [OH − ] > [H 3 O + ] are basic. [OH − ] > 1.0 × 10 −7 M Chapter 15 Section 1 Aqueous Solutions and the Concept of pH

3 Some Strong Acids and Some Weak Acids Chapter 15 Section 1 Aqueous Solutions and the Concept of pH

4 Concentrations and K w Chapter 15 Section 1 Aqueous Solutions and the Concept of pH

5 The pH Scale The pH of a solution is defined as the negative of the common logarithm of the hydronium ion concentration, [H 3 O + ]. pH = −log [H 3 O + ] example: a neutral solution has a [H 3 O + ] = 1×10 −7 The logarithm of 1×10 −7 is −7.0. pH = −log [H 3 O + ] = −log(1 × 10 −7 ) = −(−7.0) = 7.0 Chapter 15 Section 1 Aqueous Solutions and the Concept of pH

6 pH Values as Specified [H3O+] Chapter 15 Section 1 Aqueous Solutions and the Concept of pH

7 The pOH of a solution is defined as the negative of the common logarithm of the hydroxide ion concentration, [OH − ]. pOH = −log [OH – ] example: a neutral solution has a [OH – ] = 1×10 −7 The pH = 7.0. The negative logarithm of K w at 25°C is 14.0. pH + pOH = 14.0 The pH Scale Chapter 15 Section 1 Aqueous Solutions and the Concept of pH

8 The pH Scale Chapter 15 Section 1 Aqueous Solutions and the Concept of pH

9 Approximate pH Range of Common Materials Chapter 15 Section 1 Aqueous Solutions and the Concept of pH

10 pH of Strong and Weak Acids and Bases Chapter 15 Section 1 Aqueous Solutions and the Concept of pH

11 pH Values of Some Common Materials Chapter 15 Section 1 Aqueous Solutions and the Concept of pH

12 Indicators and pH Meters Acid-base indicators are compounds whose colors are sensitive to pH. Indicators change colors because they are either weak acids or weak bases. Chapter 15 Section 2 Determining pH and Titrations

13 Indicators and pH Meters The pH range over which an indicator changes color is called its transition interval. Indicators that change color at pH lower than 7 are stronger acids than the other types of indicators. They tend to ionize more than the others. Indicators that undergo transition in the higher pH range are weaker acids. Chapter 15 Section 2 Determining pH and Titrations

14 Indicators and pH Meters A pH meter determines the pH of a solution by measuring the voltage between the two electrodes that are placed in the solution. The voltage changes as the hydronium ion concentration in the solution changes. Measures pH more precisely than indicators Chapter 15 Section 2 Determining pH and Titrations

15 Color Ranges of Indicators Chapter 15 Section 2 Determining pH and Titrations

16 Chapter 15 Section 2 Determining pH and Titrations Color Ranges of Indicators

17 Chapter 15 Section 2 Determining pH and Titrations Color Ranges of Indicators

18 Titration Neutralization occurs when hydronium ions and hydroxide ions are supplied in equal numbers by reactants. H 3 O + (aq) + OH − (aq) 2H 2 O(l) Chapter 15 Section 2 Determining pH and Titrations 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.

19 Titration, continued Equivalence Point The point at which the two solutions used in a titration are present in chemically equivalent amounts is the equivalence point. The point in a titration at which an indicator changes color is called the end point of the indicator. Chapter 15 Section 2 Determining pH and Titrations

20 Titration, continued Equivalence Point, continued Indicators that undergo transition at about pH 7 are used to determine the equivalence point of strong- acid/strong base titrations. The neutralization of strong acids with strong bases produces a salt solution with a pH of 7. Chapter 15 Section 2 Determining pH and Titrations

21 Titration, continued Equivalence Point, continued Indicators that change color at pH lower than 7 are used to determine the equivalence point of strong- acid/weak-base titrations. The equivalence point of a strong-acid/weak-base titration is acidic. Chapter 15 Section 2 Determining pH and Titrations

22 Titration, continued Equivalence Point, continued Indicators that change color at pH higher than 7 are used to determine the equivalence point of weak- acid/strong-base titrations. The equivalence point of a weak-acid/strong-base titration is basic. Chapter 15 Section 2 Determining pH and Titrations

23 Titration Curve for a Strong Acid and a Strong Base Chapter 15 Section 2 Determining pH and Titrations

24 Titration Curve for a Weak Acid and a Strong Base Chapter 15 Section 2 Determining pH and Titrations

25 Molarity and Titration The solution that contains the precisely known concentration of a solute is known as a standard solution. A primary standard is a highly purified solid compound used to check the concentration of the known solution in a titration The standard solution can be used to determine the molarity of another solution by titration. Chapter 15 Section 2 Determining pH and Titrations

26 Performing a Titration, Part 1 Chapter 15 Section 2 Determining pH and Titrations

27 Performing a Titration, Part 1 Chapter 15 Section 2 Determining pH and Titrations

28 Performing a Titration, Part 1 Chapter 15 Section 2 Determining pH and Titrations

29 Performing a Titration, Part 2 Chapter 15 Section 2 Determining pH and Titrations

30 Performing a Titration, Part 2 Chapter 15 Section 2 Determining pH and Titrations

31 Performing a Titration, Part 2 Chapter 15 Section 2 Determining pH and Titrations

32 Molarity and Titration, continued To determine the molarity of an acidic solution, 10 mL HCl, by titration 1.Titrate acid with a standard base solution 20.00 mL of 5.0 × 10 −3 M NaOH was titrated 2.Write the balanced neutralization reaction equation. HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) Chapter 15 Section 2 Determining pH and Titrations 1 mol 1 mol 3.Determine the chemically equivalent amounts of HCl and NaOH.

33 Molarity and Titration, continued 4.Calculate the number of moles of NaOH used in the titration. 20.0 mL of 5.0 × 10 −3 M NaOH is needed to reach the end point Chapter 15 Section 2 Determining pH and Titrations 5.amount of HCl = mol NaOH = 1.0 × 10 −4 mol 6.Calculate the molarity of the HCl solution

34 Molarity and Titration, continued 1.Start with the balanced equation for the neutralization reaction, and determine the chemically equivalent amounts of the acid and base. 2. Determine the moles of acid (or base) from the known solution used during the titration. 3.Determine the moles of solute of the unknown solution used during the titration. 4. Determine the molarity of the unknown solution. Chapter 15 Section 2 Determining pH and Titrations

35 Molarity and Titration, continued Sample Problem F In a titration, 27.4 mL of 0.0154 M Ba(OH) 2 is added to a 20.0 mL sample of HCl solution of unknown concentration until the equivalence point is reached. What is the molarity of the acid solution? Chapter 15 Section 2 Determining pH and Titrations

36 Molarity and Titration, continued Ba(OH) 2 + 2HCl BaCl 2 + 2H 2 O 1 mol 2 mol Chapter 15 Section 2 Determining pH and Titrations Sample Problem F Solution Given: volume and concentration of known solution = 27.4 mL of 0.0154 M Ba(OH) 2 Unknown: molarity of acid solution Solution: 1.balanced neutralization equation chemically equivalent amounts

37 Molarity and Titration, continued Sample Problem F Solution, continued 2. volume of known basic solution used (mL) amount of base used (mol) Chapter 15 Section 2 Determining pH and Titrations 3. mole ratio, moles of base used moles of acid used from unknown solution

38 Molarity and Titration, continued Sample Problem F Solution, continued 4. volume of unknown, moles of solute in unknown molarity of unknown Chapter 15 Section 2 Determining pH and Titrations

39 Molarity and Titration, continued Sample Problem F Solution, continued 1.1 mol Ba(OH) 2 for every 2 mol HCl. Chapter 15 Section 2 Determining pH and Titrations 2. 3.

40 Molarity and Titration, continued Sample Problem F Solution, continued Chapter 15 Section 2 Determining pH and Titrations 4.

41 Ch. 15: Solutions / Acids and Bases Dilutions

42 Ch. 15: Solutions / Acids and Bases Concentration How to Make a Solution

43 Ch. 15: Solutions / Acids and Bases Concentration How to Make a Solution

44 Concentration Dilute small amount of solute in the solution Concentrated large amount of solute in the solution These are vague terms without definite boundaries They can be used to compare solutions

45 Molarity is most often used to specify the concentration of a solution number of moles of solute in one liter of solution that does not mean solute added to one liter of solvent- WHY? units: moles/liter = M

46 Example 1 3.7 moles of HCl is added to water to make 500. mL of solution. What is the molarity? moles are given total volume of soln is given

47 Example 2 How many liters of 0.20 M solution can be from 3.00 mol of NaCl?

48 Example 3 How many moles of NaCl are needed to make 11.0 L of 0.15 M solution? How many grams of NaCl would that be?

49 Example 4 21.0 g of NaOH is dissolved in enough water to make 500. mL of solution. What is the molarity? find moles from grams total volume of solution is given

50 How to make a solution Calculate the number of grams of solute that you need to make a certain concentration of solution Measure that amount out in a beaker Dissolve that solute with distilled water Use glass stirring rod to quicken process Pour that solution in the correct volumetric flask

51 How to Make a Solution Rinse the beaker with more distilled water into the flask Add more distilled water to the flask until the level reaches the etched line Put stopper in flask and turn the flask upside down many times to mix

52 How to Make a Solution How much solid solute would you need to dissolve to make 20. mL of 2.0 M NaOH?

53 Dilutions What are we changing when we dilute a solution? Moles or volume or both? What will happen to the molarity?

54 Diluting a Solution If the number of moles stays the same, then we can rearrange the equation for molarity

55 Example If a solution starts off at 12.0 M, what molarity will it have if 50. mL of it is diluted to 100. mL?

56 Example 2 I have a 3.7 M NaCl solution but I want 300. mL of 0.10 M solution. How many mL of the concentrated solution should I dilute?

57 Example 2 20.0 mL of a solution is diluted to 200. mL. The new concentration is 0.10 M. What was the original concentration?


Download ppt "Hydronium Ions and Hydroxide Ions Self-Ionization of Water In the self-ionization of water, two water molecules produce a hydronium ion and a hydroxide."

Similar presentations


Ads by Google