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Chapter 16: Equilibrium in Acid-Base Systems 16.1a: Self-ionization of Water K w pH and pOH.

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Presentation on theme: "Chapter 16: Equilibrium in Acid-Base Systems 16.1a: Self-ionization of Water K w pH and pOH."— Presentation transcript:

1 Chapter 16: Equilibrium in Acid-Base Systems 16.1a: Self-ionization of Water K w pH and pOH

2 Self-ionization of water  If water is a weak electrolyte, what does it ionize in to? H 2 O (l) ⇄ OH - (aq) + H + (aq) OR 2H 2 O (l) ⇄ OH - (aq) + H 3 O + (aq)  the concentrations of the ions are only 1.0 x M at 25°C  the product of these concentrations will remain constant for any solution at the same temperature

3 Self-ionization of Water  concentrations or molarities can be written with brackets  For example: concentration of A = [A] = 2.0 M Kw:Kw:Kw:Kw: the ionization constant of water the ionization constant of water the product of [OH - ] and [H + ] the product of [OH - ] and [H + ]  at 25 o C

4 KwKwKwKw  subject to the same restricition as any other equilibrium constant (T, P)  Will acidic solutions have more H + or OH - ? [H + ]>[OH - ]: acidic [H + ]>[OH - ]: acidic [OH - ]>[H + ]: basic [OH - ]>[H + ]: basic [OH - ]=[H + ]: neutral [OH - ]=[H + ]: neutral  can find the [OH - ] or [H + ] from a mole ratio of the dissociation or reaction in the water of the acid or base

5 Example 1  Calculate the H + concentration in a 1.5 M Ca(OH) 2 solution What does the Ca(OH) 2 create in solution? What does the Ca(OH) 2 create in solution? Ca(OH) 2  Ca OH - Ca(OH) 2  Ca OH - can calculate the OH - concentration can calculate the OH - concentration use K w to calculate the H + concentration use K w to calculate the H + concentration

6 Example 2  Calculate the H + and OH - concentration of a 1.0x10 -4 M solution of HNO 3 Find the [H + ] from mole ratio Find the [H + ] from mole ratio Find [OH - ] from K w and [H + ] Find [OH - ] from K w and [H + ] HNO 3  H + + NO 3 -

7 pH scale  more convenient than using concentrations  pH=-log [H + ]  pOH=-log [OH - ]  pH increases as [H + ] decreases  pH < 7: acid  pH > 7: base  pH = 7: neutral

8 pH  for any solution at 25 o C: -log[H + ] + -log[OH - ] = -log([H + ][OH - ] -log[H + ] + -log[OH - ] = -log([H + ][OH - ] = -log(1.0x ) =14 = -log(1.0x ) =14  = pH + pOH

9 Example 3  Find the pH of a 1.0x10 -3 M NaOH solution find [OH - ] using mole ratio find [OH - ] using mole ratio find [H + ] from K w find [H + ] from K w find pH from [H + ] find pH from [H + ]OR find pOH from [OH - ] find pOH from [OH - ] find pH from pOH find pH from pOH

10 Example 3 sol’n OR

11 Calculating [H + ] and [OH - ]  reverse the pH equation  The pH of a solution is Find the [H+] and [OH-] and determine whether it is acidic, basic, or neutral. [OH - ] > [H + ] so solution is basic

12 Example  A shampoo has a pH of Calculate the pOH, [H + ] and [OH - ]. Is it acidic, basic, or neutral? pH < 7 so acidic

13 Homework  Textbook p716 #1-6  p718 #7  LSM 16.1B


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