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Kw of Water

Kw of Water. Chapter 16 part III. Kw of Water. From the auto-ionization expression of water we get the equilibrium constant expression. 2 H 2 O  H 3 O + + OH- Kw= [H 3 O+ ][OH-] Why not H 2 O? Kw= [H+ ][OH-]. Kw is the ion-product. Kw is aka dissociation constant of water

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Kw of Water

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  1. Kw of Water Chapter 16 part III

  2. Kw of Water • From the auto-ionization expression of water we get the equilibrium constant expression. • 2 H2O  H3O+ + OH- • Kw= [H3O+ ][OH-] • Why not H2O? • Kw= [H+ ][OH-]

  3. Kw is the ion-product • Kw is aka dissociation constant of water • It has been found experimentally that at 25 °C, • [H+ ]=[OH-] and they both equal 1X10-7M • Since Kw= [H+ ][OH-] = [1X10-7M]2 • Then Kw = 1X 10-14.

  4. What does this mean? • This means that any aqueous solution at 25 ° C, no matter what it (the water) contains, the product of [H+] and [OH-] MUST always equal 1.0 X 10-14. • There are 3 possibilities: • A neutral solution where [H+]=[OH-]. • An acid solution where [H+]>[OH-]. • A basic solution where [H+]<[OH-].

  5. In all 3 of the situations: • Kw = [H+][OH-] = 1.0 X 10-14. • So in any given aqueous situation, one may calculate the [H+] or [OH-] as required for any solution at 25°C. • State if Acidic, Basic or Neutral. • A. 1.0X10-5 M OH- 1.0 X10-9 M H+ • B. 1.0X10-7 M OH- 1.0 X10-7 M H+ • C. 1.0X10-15 M OH- 10.0 M H+

  6. Answers • A. Basic • B. Neutral • C. Acid

  7. How to Solve for the [ions] • Kw = [H+][OH-] • 1.0 X 10-14= [H+][OH-] • [H+] = [OH-]/(1.0 X 10-14)

  8. Kw varies with Temperature • So if the Kw at 60° C is 1.0 X 10-13 • Then, using Le Châtlier’s principle predict whether the ionization of water is endo or exothermic. • 2 H2O  H3O+ + OH- • Kw increases • from 1.0 X 10-14 to 1.0 X 10-13 • Therefore the system adjusted to form more product in the presence of heat. • This indicates the reaction is endothermic

  9. Example • Calculate the [OH-] and [H+] at 60° C in a neutral solution. • Neutral means what about [H+] to [OH-]? • Answer: [H+] = [OH-] • So: Kw= [H+] x [OH-] = 1.0 X 10-13 • [H+] = [OH-] = (1.0 X 10-13 )½ • =3.0 X 10-7 M

  10. pH scale is an easy way to represent acidity. • pH = -log[H+] • At a neutral solution at 25 °C • [H+] = [OH-] = (1.0 X 10-14)½ = 1.0 X 10-7 • What is the pH of this? • pH = -log[H+] • = -log(1.0 X 10-7) • Take out you calculator and what do you get? • 7.00

  11. Sig Figs in Log problems • The number of sig figs in an original number equals the number of decimal places in the pH. • Example: • If sample is Kw= [H+] = 1.0 X 10-7 • How many sig fig? • 2 • This pH is … 7.00 2 decimal places for the two sig figs.

  12. pH vs. pOH • If pH is = -log [H+] • Then pOH = -log [OH-] • And pK = -log K • Note that pH changes by 1 for every power of 10 in the change of concentration.

  13. Examples • Calculate the pH and pOH of each • 1.0 X 10-3 M OH- • pOH = 3.00 pH = 11.00 • H+ = Kw/[OH-] = (1.0 X 10-14)/(1.0 X 10-3) • = 1.0 x 10-11 • 1.0 M H+ • pH = 0.00 [OH-]=Kw/[H+]= 1.0x10-14/1 • pOH = 14

  14. Remember • Kw =[H+][OH-] • And therefore, • -log K = -log [H+] + -log [OH-] • log K = log[H+] + log[OH-] • pKw = pH = pOH • At 25 °C pKw = 14.00 (1.0 X 10-14) • Thus pH = pOH = 14 at 25 °C

  15. Summary of General Strategies • Think Chemistry: focus on solution and the components. It is usually easy to identify one reaction that is not important. • Be systematic: Acid-base problems require step by step approach. • Be Flexible: Although all acid-base problems are similar, important differences do occur. Do not force a given problem into matching a problem you have solved before.

  16. Summary of General Strategies • Be patient: the complete solution to a complicated problem cannot be seen immediately in all its detail. Pick the problem apart into workable steps. • Understand & Think: don’t just memorize.

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