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Acids, Bases & pH

Acids, Bases & pH. Types of solutes. no conductivity. Non-electrolyte - No dissociation into ions, all molecules in solution. sugar. Types of solutes. conductivity. Electrolyte - Dissociation into ions in solution. H +. Cl -. Electrolytes. Acids, bases, and salts are electrolytes.

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Acids, Bases & pH

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  1. Acids, Bases & pH

  2. Types of solutes no conductivity Non-electrolyte - No dissociation into ions, all molecules in solution sugar

  3. Types of solutes conductivity Electrolyte - Dissociation into ions in solution. H+ Cl-

  4. Electrolytes • Acids, bases, and salts are electrolytes. • HNO3, NaOH, KBr.

  5. Acid-Base Theory - Arrhenius • Arrhenius defined an acid as follows: • An acid is a substance that produces H+ ions when dissolved in water (now described as hydronium rather than H+).

  6. Hydronium Ion (H3O+) or (H+(aq))

  7. Acid-Base Theory - Arrhenius • Arrhenius defined a base as follows: • A base is a substance that produces OH- ions when dissolved in water

  8. Acid-Base Theory - Arrhenius • We can demonstrate these definitions as follows: H2O HNO3(l) H+(aq) + NO3-(aq) H2O Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)

  9. pH scale • The pH scale is used to indicate how acidic or basic a solution is by measuring the relative amount of H+ and OH- in a solution.

  10. The pH scale

  11. pH and pOH scales Acidic Neutral Basic pH + pOH = 14.00

  12. pH and pOH formulas • Remember that H+ = H3O+ • pH = -log[H+] • pOH = -log[OH-] • [H+] x [OH-] = 1.0 x 10-14 • pH + pOH = 14.00 • [H+] = antilog(-pH) • [OH-] = antilog(-pOH)

  13. Find the [OH-] in a solution where [H3O+] = 6.80 x 10-10. • pH = -log[H+] • pOH = -log[OH-] • [H+] x [OH-] = 1.0 x 10-14 • pH + pOH = 14.00 • [H+] = antilog(-pH) • [OH-] = antilog(-pOH)

  14. Find the [OH-] in a solution where [H3O+] = 6.80 x 10-10. • [H+] x [OH-] = 1.0 x 10-14 • [6.80 x 10-10] x [OH-] = 1.0 x 10-14 1.47 x 10-5 M

  15. Find the pH of a solution where [H3O+] = 7.01 x 10-6. • pH = -log[H+] • pOH = -log[OH-] • [H+] x [OH-] = 1.0 x 10-14 • pH + pOH = 14.00 • [H+] = antilog(-pH) • [OH-] = antilog(-pOH)

  16. Find the pH of a solution where [H3O+] = 7.01 x 10-6. • pH = -log[H+] • pH = -log[7.01 x 10-6] 5.15

  17. Find the [OH-] in a solution with a pOH of 4.976. • pH = -log[H+] • pOH = -log[OH-] • [H+] x [OH-] = 1.0 x 10-14 • pH + pOH = 14.00 • [H+] = antilog(-pH) • [OH-] = antilog(-pOH)

  18. Find the [OH-] in a solution with a pOH of 4.976. [OH-] = antilog(-pOH) [OH-] = antilog(-4.976) 1.06 x 10-5 M

  19. What is the pH of a solution with a [OH-] = 9.80 x 10-9. • pH = -log[H+] • pOH = -log[OH-] • [H+] x [OH-] = 1.0 x 10-14 • pH + pOH = 14.00 • [H+] = antilog(-pH) • [OH-] = antilog(-pOH)

  20. What is the pH of a solution with a [OH-] = 9.80 x 10-9. pOH = -log[OH-] pOH = -log[9.80 x 10-9] pOH = 8.01 pH + pOH = 14.00 14.00 – 8.01 = 5.99

  21. Acid Base Neutralization • Acid+ Base  • HCl + NaOH • HCl + NaOHNaCl + HOH • Acid + Base  Salt + Water salt water

  22. Acid-Base Neutralization Reactions HCl + NaOH  NaCl + H2O • The products of an acid-base neutralization reaction are a salt made up of the positive ion of the base (Na+) and the negative ion (Cl-) of the acid, and water.

  23. HCl + NaOHNaCl + H2O • Remember, in writing the correct formula for the salt, the total positive charge of the ions in the salt must equal the total negative charge of the ions in the salt. • Make sure you write the correct formula for the products, salt and water first, then balance the equation.

  24. Write the balanced equation for the acid-base neutralization of perchloric acid, HClO4 and potassium hydroxide, KOH HClO4 + KOH  KClO4 + H2O

  25. Write the balanced equation for the acid-base neutralization of nitric acid, HNO3 and calcium hydroxide Ca(OH)2 Ca(OH)2 + 2HNO3 Ca(NO3)2 + 2H2O

  26. MA x VA = MB x VB • Acids and bases often react together in what is called a neutralization reaction. The amount (volume) of acid or base and the strength (molarity) of acid and base needed to neutralize each other can “sometimes” be found by using the formula above.

  27. MA x VA = MB x VB • We can use this formula when the acid and base react together in a 1:1 ratio. Formula can be used: HClO4 + KOH  KClO4 + H2O Formula cannot be used: Ca(OH)2 + 2HNO3Ca(NO3)2 + 2H2O

  28. MA x VA = MB x VB • 15.6ml of acid neutralize 25.8ml of 0.50M base. What is the molarity of the acid? 0.83M

  29. MA x VA = MB x VB • 32.6 ml of 1.25M acid neutralize how many ml of 5.60M base? 7.28 mL

  30. Homework • Summarize the Lab: “Acid – Base Titration” • Worksheet: Acids, Bases and pH

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