pH at the equivalence point a Total volume b Neutralization rxn HNO 2 aq NaOH

Ph at the equivalence point a total volume b

This preview shows page 8 - 12 out of 16 pages.

3.pH at the equivalence point:a.Total volume:_______________________________________ 2 O (ℓ) (aq) 2 O (ℓ) = pK a (aq) 8
Image of page 8
D. Acid-Base Indicators 1. An indicator is a weak acid or a weak base that changes color at the equivalence point "indicating" that the reaction is complete. 2. We will use a weak acid as our example. General formula: HIn HIn (aq) + H 2 O(ℓ) H 3 O + (aq) + In (aq) green purple non-ionized ionized a. In acid, the equilibrium" lies to the left", the color of the indicator is green. b. In base, the equilibrium "lies to the right", the color of the indicator is purple. c. Generally, if [HIn] 10, non-ionized form predominates (green) [In ] [HIn] 0.10, ionized form predominates (purple) [In ] [HIn] [In ] the indicator color is the combination of the two (green + purple) 3. Phenolpthalein is used in the titration of a strong acid with a strong base. (pH 8.3-10) in acid - colorless in base - pink 4. The indicator should change color in the vertical region of the titration curve. 5. Methyl red is an appropriate indicator for the titration of a weak acid with a strong base. (pH 4.2-6.3) 9
Image of page 9
IV. Solubility Equilibria A. Solubility product, K sp 1. The equilibrium constant for a slightly soluble salt is the K sp , called the solubility product constant. Because the concentration of a solid is constant, it does not appear in the equilibrium constant expression. a. AgCl (s) Ag + (aq) + Cl (aq) K sp = [Ag + ][Cl ] = 1.6 x 10 10 b. Ca 3 (PO 4 ) 2 (s) 3 Ca 2+ (aq) + 2 PO 4 3 (aq) K sp = [Ca 2+ ] 3 [PO 4 3 ] 2 = 1.3 x 10 32 2. Table 17.4, page 733 in the textbooks lists the solubility product constants for several slightly soluble salts at 25 C. Please refer to this table for K sp values. B. Calculations involving K sp and Solubility Molar Solubility and Solubility 1. Molar solubility = number of moles of solute = mol = M liter of solution L 2. Solubility = number of grams of solute = g or g volume of solution 100mL L Sample calculations: 1. The solubility of lead chromate, PbCrO 4 , is 4.5 x 10 5 g/L. Calculate the solubility product constant, K sp , for lead chromate. The molar mass of lead chromate is 323 g/mol. PbCrO 4 (s) Pb 2+ (aq) + CrO 4 2 (aq) 10
Image of page 10
Image of page 11
Image of page 12

You've reached the end of your free preview.

Want to read all 16 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture