1. Part I: Common Ion Effect

2.  When the salt with the anion of a weak acid is added to that acid,  It reverses the dissociation of the acid.  Lowers the percent dissociation of the acid.  The same principle applies to salts with the cation of a weak base.  The calculations are the same as last chapter.

3.  A solution that resists a change in pH.  Either a weak acid and its salt or a weak base and its salt.  We can make a buffer of any pH by varying the concentrations of these solutions.  Same calculations as before.  Calculate the pH of a solution that is.50 M HC 2 H 3 O 2 and.25 M NaC 2 H 3 O 2 (Ka = 1.8 x 10 -5 )

4. Na + is a spectator and the reaction we are worried about is R HC 2 H 3 O 2 H + + C 2 H 3 O 2 - x xx -x 0.50-x0.25+x I 0.50 M 0 0.25 M E C

5.  Do the math  Ka = 1.8 x 10 -5 1.8 x 10 -5 = x (0.25+x) (0.50-x) = x (0.25) (0.50) x = 3.6 x 10 -5 pH = -log (3.6 x 10 -5 ) = 4.44 HC 2 H 3 O 2 H + + C 2 H 3 O 2 -

6.  Ka = [H + ] [A - ] [HA]  so [H + ] = Ka [HA] [A - ]  The [H + ] depends on the ratio [HA]/[A - ]  taking the negative log of both sides  pH = -log(Ka [HA]/[A - ])  pH = -log(Ka)-log([HA]/[A - ])  pH = pKa + log([A - ]/[HA])

7.  pH = pKa + log(base/acid)  Works for an acid and its salt  Like HNO 2 and NaNO 2  Or a base and its salt  Like NH 3 and NH 4 Cl  But remember to change K b to K a

8. pH = pKa + log [base]/[acid] pH = -log (1.8 x 10 -5 ) + log (.25/.50) pH = 4.44

9. Let’s take the solution and add 0.01mol of NaOH to it. 1. Write the net ionic equation for the reaction of HC 2 H 3 O 2 + NaOH HC 2 H 3 O 2 + OH - C 2 H 3 O 2 - + H 2 O 2. Do the stoichiometry using moles (Use a BRA Chart)

10. HC 2 H 3 O 2 + OH - C 2 H 3 O 2 - + H 2 O B.50mol.01mol.25mol R -.01 -.01 +.01 A.49 mol 0.26mol *Assume 1L of solution and we now have our starting concentrations

11. I.49 0.26 C -x +x +x E.49 –x x.26+x K a = 1.8 x10 -5 = x(.26+x) / (.49-x) 5% rule 1.8 x10 -5 = x(.26) / (.49) x = 3.4 x 10 -5 pH = 4.47 R HC 2 H 3 O 2 H + + C 2 H 3 O 2 -

12. pH = pKa + log [base]/[acid] pH = -log (1.8 x 10 -5 ) + log (.26/.49) pH = 4.47

13.  If we had added 0.010 mol of NaOH to 1 L of water, what would the pH have been.  0.010 M OH - NaOH Na + + OH -.010.010.010  pOH = -log(0.010) = 2.00  pH = 12.00  But with a mixture of an acid and its conjugate base the pH doesn’t change much  Called a buffer.

14. Calculate the pH 0.75 M lactic acid (HC 3 H 5 O 3 ) and 0.25 M sodium lactate (Ka = 1.4 x 10 -4 ) pH = pKa + log(base/acid) pH = -log(1.4 x 10 -4 ) + log (.25/.75) pH = 3.38

15. Calculate the pH  0.25 M NH 3 and 0.40 M NH 4 Cl (K b = 1.8 x 10 -5 )  Ka = 5.6 x 10 -10  remember its the ratio base over acid pH = -log (5.6 x 10 -10 ) + log(.25/.40) pH = 9.05

16.  What would the pH be if.020 mol of HCl is added to the lactic acid solution?  0.75 M lactic acid (HC 3 H 5 O 3 ) and 0.25 M sodium lactate (Ka = 1.4 x 10 -4 ) HC 3 H 5 O 3 H + + C 3 H 5 O 3 - Before 0.75 mol +.020.25 mol Reactions +.02mol -.02 -.02 After 0.77 mol00.23 mol

17. pH = pKa + log (base/acid) pH = -log(1.4 x 10 -4 ) + log (.23/.77) pH = 3.33, compared to 3.38 before adding the HCl

18.  What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L of the solutions.  0.75 M lactic acid (HC 3 H 5 O 3 ) and 0.25 M sodium lactate (Ka = 1.4 x 10 -4 ) HC 3 H 5 O 3 + OH - H 2 O + C 3 H 5 O 3 - B 0.75 mol.05 mol 0.25 mol R -.05 -.05 +.05 A 0.70 mol 0 mol 0.30 mol

19. pH = pKa + log (base/acid) pH = -log(1.4 x 10 -4 ) + log (.30/.70) pH = 3.49, compared to 3.38 before adding the NaOH

20.  What would the pH be if.020 mol of HCl is added to 1.0 L of ammonia solution.  0.25 M NH 3 and 0.40 M NH 4 Cl K b = 1.8 x 10 -5 so Ka = 5.6 x 10 -10 NH 3 + H + NH 4 + B 0.25 mol.02 mol0.40 mol R -.02 -.02 +.02 A 0.23 mol0 mol0.42 mol Net Ionic

21. pH = pKa + log (base/acid) pH = -log(5.6 x 10 -10 ) + log (.23/.42) pH = 8.99, compared to 9.05 before adding the HCl

22.  What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L of ammonia solution.  0.25 M NH 3 and 0.40 M NH 4 Cl K b = 1.8 x 10 -5 so Ka = 5.6 x 10 -10 NH 3 + H 2 O NH 4 + + OH - B 0.40 mol 0.25 mol.050mol R +.05 -.05 -.05 A 0.45 mol 0.20 mol 0 mol

23. pH = pKa + log (base/acid) pH = -log(5.6 x 10 -10 ) + log (.45/.20) pH = 9.60, compared to 9.05 before adding the NaOH

24.  The pH of a buffered solution is determined by the ratio [A - ]/[HA].  As long as this doesn’t change much the pH won’t change much.  The more concentrated these two are the more H + and OH - the solution will be able to absorb.  Larger concentrations = bigger buffer capacity.

25.  Calculate the change in pH that occurs when 0.020 mol of HCl(g) is added to 1.0 L of each of the following:  5.00 M HAc and 5.00 M NaAc  0.050 M HAc and 0.050 M NaAc  Ka= 1.8x10 -5  pH = pKa

26.  Calculate the change in pH that occurs when 0.040 mol of HCl(g) is added to 1.0 L of 5.00 M HAc and 5.00 M NaAc  Ka= 1.8x10 -5 HAc H + + Ac - B 5.00 mol.04mol5.00 mol R +.04 -.04 -/04 A 5.04 mol4.96 mol pH = 4.74 Compared to 4.74 before acid was added

27.  Calculate the change in pH that occurs when 0.040 mol of HCl(g) is added to 1.0 L of 0.050 M HAc and 0.050 M NaAc  Ka= 1.8x10 -5 HAc H + + Ac - B 0.050 mol.04mol 0.050 mol R +.04 -.04 -.04 A 0.090 mol 0 0.010 mol pH = 3.79 Compared to 4.74 before acid was added

28.  The best buffers have a ratio [A - ]/[HA] = 1  This is most resistant to change  True when [A - ] = [HA]  Makes pH = pKa (since log 1 = 0)