1. Acids and Bases II Dr. Ron Rusay Summer 2004 © Copyright 2004 R.J. Rusay

2. Water as an Acid and a Base  Amphoteric substances can act as either an acid or a base Water acting as an acid:Water acting as an acid: NH 3 + H 2 O  NH 4 +1 + OH -1 NH 3 + H 2 O  NH 4 +1 + OH -1 Water acting as a base:Water acting as a base: HCl + H 2 O  H 3 O +1 + Cl -1 HCl + H 2 O  H 3 O +1 + Cl -1 Water reacting with itself as both:Water reacting with itself as both: H 2 O + H 2 O  H 3 O +1 + OH -1

3. Water as an Acid and a Base  Water is amphoteric. It can behave either as an acid or a base. H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH  (aq) conj conj conj conj acid 1 base 1 acid 2 base 2 acid 1 base 1 acid 2 base 2  The equilibrium expression for pure water is: K w = [H 3 O + (aq) ] [OH  (aq) ]

4. Water: Self-ionization

5. Autoionization of Water  Water is an extremely weak electrolyte therefore are only a few ions present: K w = [H 3 O +1 ] [OH -1 ] = 1 x 10 -14 @ 25°C NOTE: the concentration of H 3 O +1 and OH -1 are equalNOTE: the concentration of H 3 O +1 and OH -1 are equal [H 3 O +1 ] = [OH -1 ] = 10 -7 M @ 25°C[H 3 O +1 ] = [OH -1 ] = 10 -7 M @ 25°C K w is called the ion product constant for water: as [H 3 O +1 ] increases, [OH - ] decreases and vice versa.K w is called the ion product constant for water: as [H 3 O +1 ] increases, [OH - ] decreases and vice versa.

6. Acidic and Basic Solutions  Acidic solutions have: a larger [H +1 ] than [OH -1 ]  Basic solutions have: a larger [OH -1 ] than [H +1 ]  Neutral solutions have [H +1 ] = [OH -1 ] = 1 x 10 -7 M [H +1 ] = 1 x 10 -14 [OH -1 ] [OH -1 ] = 1 x 10 -14 [H +1 ]

7. The pH Scale  pH   log [H + ]  log [H 3 O + ]  1 pH unit corresponds to a factor of 10  pH in water ranges from 0 to 14. K w = 1.00  10  14 = [H + ] [OH  ] -log K w = -log [H 3 O +1 ] -log [OH -1 ] -log K w = -log [H 3 O +1 ] -log [OH -1 ] pK w = pH + pOH = 14.00  As pH rises, pOH falls (Sum = 14.00).

8. pH & pOH  pH = -log[H 3 O +1 ]pOH = -log[OH -1 ] pH water = -log[10 -7 ] = 7 = pOH waterpH water = -log[10 -7 ] = 7 = pOH water  [H +1 ] = 10 -pH [OH -1 ] = 10 -pOH  pH 7 is basic, pH = 7 is neutral  The lower the pH, the more acidic the solution; The higher the pH, the more basic the solution.  1 pH unit corresponds to a factor of 10  pOH = 14 - pH

9. There are no theoretical limits on the values of pH or pOH. (e.g. pH of 2.0 M HCl is -0.301, the pH at Iron Mountain, California is ~ -2 to -3)

10. What’s in these household products? Acids or bases? Strong or weak? Should you be concerned about safety?

11. The pH of Some Familiar Aqueous Solutions [H 3 O + ] [OH - ] [OH - ] = KWKW [H 3 O + ] neutral solution acidic solution basic solution [H 3 O + ]> [OH - ] [H 3 O + ]< [OH - ] [H 3 O + ] = [OH - ] What’s your diet? Your urine will tell!

12. The pH Scale What is the pH of 6M hydrochloric acid?

13. Example #1  Determine the given information and the information you need to find Given [H +1 ] = 10.0 MFind [OH -1 ]  Solve the Equation for the Unknown Amount Determine the [H +1 ] and [OH -1 ] in a 10.0 M H +1 solution

14.  Convert all the information to Scientific Notation and Plug the given information into the equation. Given [H +1 ] = 10.0 M= 1.00 x 10 1 M K w = 1.0 x 10 -14 Example #1 (continued) Determine the [H +1 ] and [OH -1 ] in a 10.0 M H +1 solution

15. Example #2  Find the concentration of [H +1 ] Calculate the pH of a solution with a [OH -1 ] = 1.0 x 10 -6 M

16.  Enter the [H +1 ] concentration into your calculator and press the log key log(1.0 x 10 -8 ) = -8.0  Change the sign to get the pH pH = -(-8.0) = 8.0 Example #2 continued Calculate the pH of a solution with a [OH -1 ] = 1.0 x 10 -6 M

17.  Enter the [H +1 ] or [OH -1 ]concentration into your calculator and press the log key log(1.0 x 10 -3 ) = -3.0  Change the sign to get the pH or pH pOH = -(-3) = 3.0  Subtract the calculated pH or pOH from 14.00 to get the other value pH = 14.00 – 3.0 = 11.0 Calculate the pH and pOH of a solution with a [OH -1 ] = 1.0 x 10 -3 M Example #3

18.  If you want to calculate [OH -1 ] use pOH, if you want [H +1 ] use pH. It may be necessary to convert one to the other using 14 = [H +1 ] + [OH -1 ] pOH = 14.00 – 7.41 = 6.59  Enter the pH or pOH concentration into your calculator  Change the sign of the pH or pOH -pOH = -(6.59)  Press the button(s) on you calculator to take the inverse log or 10 x [OH -1 ] = 10 -6.59 = 2.6 x 10 -7 Example #4 Calculate the [OH -1 ] of a solution with a pH of 7.41

19. Calculating the pH of a Strong, Monoprotic Acid  A strong acid will dissociate 100% HA  H +1 + A -1  Therefore the molarity of H +1 ions will be the same as the molarity of the acid  Once the H +1 molarity is determined, the pH can be determined pH = -log[H +1 ]

20. Example #5  Determine the [H +1 ] from the acid concentration HNO 3  H +1 + NO 3 -1 0.10 M HNO 3 = 0.10 M H +1  Enter the [H +1 ] concentration into your calculator and press the log key log(0.10) = -1.00  Change the sign to get the pH pH = -(-1.00) = 1.00 Calculate the pH of a 0.10 M HNO 3 solution

21. Buffered Solutions  Buffered Solutions resist change in pH when an acid or base is added to it.  Used when need to maintain a certain pH in the system, eg. Blood.  A buffer solution contains a weak acid and its conjugate base  Buffers work by reacting with added H +1 or OH -1 ions so they do not accumulate and change the pH.  Buffers will only work as long as there is sufficient weak acid and conjugate base molecules present.

23. Buffers