1. Acids & Bases Lesson 9 pH, pOH, Kb for Weak Bases

2. Weak Bases calculations Helpful Hints: -Weak bases do NOT ionize 100% -There is an equilibrium state -NEED….ICE tables! -Kb can’t be used directly from table…use -Kb = kw/Ka

3. You can calculate the Kb using the Ka from the table and the equation below. KaxKb=1.0x 10 -14 @ 25 o C Kb=Kw Ka(conjugate) or Ka=Kw Kb(conjugate)

4. Weak Bases calculations Two types of questions -calculate pOH, or [OH - ] -calculate Kb

5. 1.Calculate the pH of 0.50 M NH 3. Weak BaseNeed Kb(NH 3 ) = Kw Ka(NH 4 + ) =1 x 10 -14 5.6 x 10 -10 =1.786 x 10 -5 NH 3 + H 2 O ⇄ NH 4 + +OH - I0.50 M00 C-x xx E0.50 - x xx Small Kb 0

6. Kb=[NH 4 + ][OH - ] [NH 3 ] =x 2= 1.786 x 10 -5 0.50 x = 0.002988 M= [OH - ] pOH = 2.52pOH= -Log[OH - ] pH + pOH = pKw = 14.000 pH = 11.48

7. 2.Calculate the pH of 0.20 M Na 2 CO 3. Weak BaseNeed Kb(CO 3 2- ) = Kw Ka(HCO 3 - ) =1 x 10 -14 5.6 x 10 -11 =1.786 x 10 -4 CO 3 2- + H 2 O ⇄ HCO 3 - +OH - I0.20 M00 C-xxx E0.20 - xxx Small Kb 0

8. Kb=[HCO 3 - ][OH - ] [CO 3 2- ] =x 2 = 1.786 x 10 -4 0.20 x = 0.005976 M pOH = 2.22 pH = 11.78

9. TRY: calculate [H + ], [OH - ], pH, pOH for a 0.20 M solution of NH 3.

11. 3.Calculate the pH of 0.20 M NaCl. NaCl is a neutral salt pH=7.00

12. What if asked to calculate K b ? K b indicates weak bases, then use ICE tables. need to know [ ]’s to plug in the equation.

13. EX:If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - pH =11.427pOH = 2.573 [OH - ] = 10 -2.573 [OH - ]= 0.002673 M at equilibrium line!!! [NH 4 + ][OH - ] [NH 3 ] 0.002673 0.002673 0.40 00 - 0.002673 0.3973 ICEICE (0.002673) 2 0.3973 = 1.8 x 10 -5 Kb ==

15. TOGETHER: The pOH of a 0.50 M solution of the weak acid HA is 10.64. what is K b for A - ?

17. TRY: A 0.600 M solution of the weak base hydroxylamine, NH 2 OH, has a pH of 9.904. What is K a for NH 3 OH + ?

18. Next question Strong acid and a strong base Calculate excess No ICE tables

19. 4.15.0 mL of 0.20 M HNO 3 reacts with 40.0 mL of 0.20 M KOH, calculate the pH of the resulting solution.

20. Next question It says saturated solution = equilibrium This is a solubility equilibrium- no ICE Remember, unit 3!

21. 4.Calculate the pH of a saturated solution of Mg(OH) 2. This is a solubility equilibrium- no ICE Mg(OH) 2(s) ⇄ Mg 2+ +2OH - ss2s Ksp=[Mg 2+ ]][OH - ] 2 = 5.6 x 10 -12 [s][2s] 2 = 5.6 x 10 -12 4s 3 = 5.6 x 10 -12 s=1.119 x 10 -4 M

22. 2s = [OH - ]=2.237 x 10 -4 M pOH=3.65 pH=10.35

23. Be careful Are they strong or weak acids and bases? calculations differ! Use ICE tables for only weak combinations. Is it a Ksp question? Always understand the question BEFORE answering it. Remember ALL the equations!

24. pHEquations You must know the following equations, which are all based on the ionization of water at 25 0 C! H 2 O ⇄ H + +OH - Kw=[H + ][ OH - ]=1.00 x 10 -14 pH=-Log[H + ]OrpOH=-Log[OH - ] [H + ]=10 -pH [OH - ]=10 -pOH pH+pOH=pKw=14.000

25. Homework p.153 84, 86, 87, 88, p.154 90, 91, 93.