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Chapter 15a Neutral, Acidic, and Basic Solutions

Chapter 15a Neutral, Acidic, and Basic Solutions. Because the hydronium ion and hydroxide ion concentrations are the same in pure water, it is neutral. Any solution in which [ H 3 O+ ] = [ OH− ] is also neutral. Acids increase the concentration of H 3 O+ in aqueous solutions.

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Chapter 15a Neutral, Acidic, and Basic Solutions

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  1. Chapter 15aNeutral, Acidic, and Basic Solutions • Because the hydronium ion and hydroxide ion concentrations are the same in pure water, it is neutral. • Any solution in which [H3O+] = [OH−] is also neutral.

  2. Acids increase the concentration of H3O+ in aqueous solutions. • Solutions in which the [H3O+] is greater than the [OH−] are acidic.

  3. Bases increase the concentration of OH− in aqueous solutions. • In basic solutions, the [OH−] is greater than the [H3O+].

  4. Addition of dry ice, carbon dioxide, to waterincreases the [H3O+], which is shown by the color change of the indicator bromthymol blue to yellow. The white mist is formed by condensation of water droplets because the dry ice is cold. (b) Addition of sodium peroxide to water increases the [OH−],which is shown by the color change of the indicator phenolphthalein to pink.

  5. The [H3O+] and the [OH−] of a neutral solution at 25°C both equal 1.0 × 10−7 M. • If the [H3O+] is increased to greater than 1.0 × 10−7 M, the solution is acidic. • EX: 10-6, 10-5, 10-4 • A solution containing 1.0 × 10−5 mol H3O+ ion/L at 25°C is acidic because 1.0 × 10−5 is greater than 1.0 × 10−7.

  6. If the [OH−] is increased to greater than 1.0 × 10−7 M, the solution is basic. • EX: 10-6, 10-5, 10-4 • A solution containing 1.0 × 10−4 mol OH− ions/L at 25°C is basic because 1.0 × 10−4 is greater than 1.0 × 10−7.

  7. Calculating [H3O+] and [OH–] • Strong acids and bases are considered completely ionized or dissociated in weak aqueous solutions.

  8. 1 mol of NaOH will yield 1 mol of OH− in an aqueous solution. • Therefore, a 1.0 × 10−2 M NaOH solution has an [OH−] of 1.0 × 10−2 M, as shown by the following. Notice that the [OH−] is greater than 1.0 × 10−7 M. This solution is basic.

  9. Using one concentration to find another [H3O+] and [OH-] If you know the concentration of either hydronium ions or hydroxide ions, you can determine the other. Use this formula. [OH-] x [H3O+] = 1.0 x 10-14 Rearrange to solve for your unknown.

  10. 1.0 × 10−2 M NaOH solution has an [OH−] of 1.0 × 10−2 M [OH-] x [H3O+] = 1.0 x 10-14 So, [H3O+] = 1.0 x 10-14 [OH-] Or, [H3O+] = 1.0 x 10-14 1.0 × 10−2

  11. Lets Do Some Exciting Calculations!!!!!     

  12. Determine the [H3O+] and [OH-] in a 0.01M solution of HCLO4.

  13. An aqueous solution of Ba(OH)2 has a [H3O+] of 1 x 10 -11 M. What is the [OH-]? [OH-] x [H3O+] = 1.0 x 10-14 [OH-]= 1.0 x 10-14 [H3O+] [OH-]= 1.0 x 10-14 1 x 10 -11 [OH-] =1x10-3 M

  14. An aqueous solution of Ba(OH)2 has a [H3O+] of 1 x 10 -11 M. What is the molarity of the solution? [OH-] =1x10-3 M

  15. Chapter 15.1b The pH Scale

  16. Expressing acidity or basicity in terms of the concentration of H3O+ or OH− can be cumbersome because the values tend to be very small. A more convenient quantity, called pH, also indicates the hydronium ion concentration of a solution. The letters pH stand for the French words pouvoir hydrogène, meaning “hydrogen power.”

  17. The pH of a solution is defined as the negative of the common logarithm of the hydronium ion concentration, [H3O+]. The pH is expressed by the following equation.

  18. The common logarithm of a number is the power to which 10 must be raised to equal the number. • A neutral solution at 25°C has a [H3O+] of 1 × 10−7 M. The logarithm of 1 × 10−7 is −7.0.The pH is determined as follows

  19. The pOH of a solution is defined as the negative of the common logarithm of the hydroxide ion concentration, [OH−]. A neutral solution at 25°C has a [OH−] of 1 × 10−7 M. Therefore, the pOH is 7.0.

  20. Suppose the [H3O+] in a solution is greater than the [OH−], as is true for acidic solutions. • For example, the pH of an acidic solution at 25°C with a [H3O+] of 1 × 10−6 M is 6.0. The pH of this solution is less than 7.

  21. Let’s do some calculations involving pH!!!!  Hint: If either the [H3O+] or pH is known, the other can be calculated.

  22. What is the pH of a solution if the [H3O+] is 3.4 x 10 -5 M? pH = -log [H3O+] = -log 3.4 x 10 -5 = -(-4.47) = ph 4.47

  23. What is the pH of a solution whose [H3O+] = 6.2 x 10-9 M? • pH = -log [H3O+] • = -log 6.2 x 10 -9 • = -(-8.21) = ph 8.21

  24. Determine the pH of a 0.00074 M solution of NaOH. • pOH = - log [OH-] • = - log 7.4 x 10-4 • = - (-3.13) = pOH 3.13 pH + pOH = 14 pH = 10.87

  25. Homework 15.1 • Page 523 #1-16 • Additional Practice Text page 905 # 417-423

  26. Homework 15.1 • Text page 905 # 417-423

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