1 EQUILBRIUM OF Acids and Bases Chapter 17

2 Water H 2 O can function as both an ACID and a BASE. Equilibrium constant for water = K w K w = [H 3 O + ] [OH - ] = 1.00 x at 25 o C AUTOIONIZATION

3 Water K w = [H 3 O + ] [OH - ] = 1.00 x at 25 o C In a neutral solution [H 3 O + ] = [OH - ] so [H 3 O + ] = [OH - ] = 1.00 x M Autoionization

4 [H 3 O + ], [OH - ] and pH A common way to express acidity and basicity is with pH pH = - log [H 3 O + ] In a neutral solution, [H 3 O + ] = [OH - ] = 1.00 x at 25 o C pH = -log (1.00 x ) = - (-7) = 7

5 [H 3 O + ], [OH - ] and pH What is the pH of the M NaOH solution? [H 3 O + ] = 1.0 x M pH = - log (1.0 x ) = General conclusion — Basic solution pH > 7 Neutral pH = 7 Acidic solutionpH < 7

6 [H 3 O + ], [OH - ] and pH If the pH of Coke is 3.12, it is ________. log [H 3 O + ] = - pH Take antilog and get [H 3 O + ] = 10 -pH [H 3 O + ] = = 7.6 x M

7 pX Scales In general pX = -log X pOH = - log [OH - ] pH = - log [H + ] pK w = 14 = pH + pOH

8 Weak Base Step 3. Calculate pH [OH - ] = 4.2 x M so pOH = - log [OH - ] = 3.37 Because pH + pOH = 14, pH = 10.63

9 Equilibria Involving Weak Acids and Bases Aspirin is a good example of a weak acid, K a = 3.2 x 10 -4

10 Weak Acids and Bases AcidConjugate Base acetic, CH 3 CO 2 HCH 3 CO 2 -, acetate ammonium, NH 4 + NH 3, ammonia bicarbonate, HCO 3 - CO 3 2-, carbonate A weak acid (or base) is one that ionizes to a VERY small extent (< 5%).

11 Weak Acids and Bases acetic acid, CH 3 CO 2 H (HOAc) HOAc + H 2 O  H 3 O + + OAc - Acid Conj. base (K is designated K a for ACID) [H 3 O + ] and [OAc - ] are SMALL, K a << 1.

12 Equilibrium Constants for Weak Acids Weak acid has K a < 1 Leads to small [H 3 O + ] and a pH of 2 - 7

13 Calculations with Equilibrium Constants pH of an acetic acid solution. What are your observations? pH of an acetic acid solution. What are your observations? M M 0.06 M 2.0 M a pH meter, Screen 17.9

14 Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the pH. And the equilibrium concs. of EACH Step 1. ICE table. [HOAc][H 3 O + ][OAc - ] I C E

15 Equilibria Involving A Weak Acid [HOAc] [H 3 O + ] [OAc - ] I C -x +x +x E 1.00-x x x Note that we neglect [H 3 O + ] from H 2 O.

16 Equilibria Involving A Weak Acid Step 2. Write K a expression This is a quadratic. Use quadratic formula or method of approximations (see Appendix A). HOWEVER

17 Equilibria Involving A Weak Acid Assume x is very small because K a is so small. Now we can more easily solve this approximate expression.

18 Equilibria Involving A Weak Acid Step 3. Solve K a approximate expression x = [ H 3 O + ] = [ OAc - ] = [K a 1.00] 1/2 x = [ H 3 O + ] = [ OAc - ] = 4.2 x M pH = -log (4.2 x ) = 2.37

19 Equilibria Involving A Weak Acid For many weak acids [H 3 O + ] = [conj. base] = [K a C o ] 1/2 where C 0 = initial conc. of acid Useful Rule of Thumb: If 100K a < C o, then [H 3 O + ] = [K a C o ] 1/2

20 Equilibria Involving A Weak Acid Calculate the pH of a M solution of formic acid, HCO 2 H. HCO 2 H + H 2 O  HCO H 3 O + K a = 1.8 x Approximate solution [H 3 O + ] = [K a C o ] 1/2 = 4.2 x M, pH = 3.4 Exact Solution [H 3 O + ] = [HCO 2 - ] = 3.4 x M [HCO 2 H] = x = M pH = 3.5

Weak Bases

22 Equilibrium Constants for Weak Bases Weak base has K b < 1 Leads to small [OH - ] and a pH of

23 Equilibria Involving A Weak Base You have M NH 3. Calc. the pH. NH 3 + H 2 O  NH OH - K b = 1.8 x Step 1; ICE table [NH 3 ][NH 4 + ][OH - ] I C E

24 Weak Base Step 1. ICE table [NH 3 ][NH 4 + ][OH - ] I C -x +x +x E x x x

25 Weak Base Step 2. Solve the equilibrium expression Assume x is small (100K b < C o ), so x = [OH - ] = [NH 4 + ] = 4.2 x M [NH 3 ] = x ≈ M The approximation is valid!

26 AcidsAcids ConjugateBasesConjugateBases

27 Relation of K a, K b, [H 3 O + ] and pH

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