1. (1 pt.)  Which of the following will make a buffer 
	
solution with a pH above 7?


a. 1 M NaOH and 1 M NaCl

b. 1 M HC2H3O2 and 1 M NaC2H3O2

c. 1 M HCl and 1 M NaCl
	
d. 1 M NH4Cl and 1 M NH3


2. (6 pts.)  Calculate the pH at the equivalence point for 
	the titration of 75.0 mL of 0.200 M HCN with 0.300 M KOH.  
	Ka = 4.0 x 10-10























3. (4 pts.)  What is the pH of a buffer solution containing 
	0.800 M HF and 0.500 M NaF?  Ka = 4.0 x 10-4










4. (3 pts.)  a.	What is the solubility of Mg(OH)2 in water?
		Ksp = 1.2 x 10-11














(2 pts.)  b. What is the solubility of Mg(OH)2 in 0.10 M NaOH?
             Ksp = 1.2 x 10-11












5. (4 pts.)  Calculate the [H+] of a solution prepared by 

	mixing 60.0 mL of 2.50 M KOH and 40.0 mL of 1.75 M H2SO4.










PRACTICE QUIZ KEY

1. (1 pt.)  Which of the following will make a buffer solution 
	with a pH above 7?

a. 1 M NaOH and 1 M NaCl	c. 1 M HCl and 1 M NaCl

b. 1 M HC2H3O2 and 1 MNaC2H3O2	(d.) 1 M NH4Cl and 1 M NH3

2. (6 pts.)  Calculate the pH at the equivalence point for the 
	titration of 75.0 mL of 0.200 M HCN with 0.300 M KOH.  
	Ka = 4.0 x 10-10		mol HCN = M x V

Step 1:  OH-  +  HCN ---->  H2O + CN- 	0.200 mol x 0.0750 L =
							      L
Init.   0.0150   0.0150           0			0.0150 mol
RXN    -0.0150  -0.0150         +0.0150
Final      0        0           0.0150 mol	 0.0150 mol 
					      total volume  
(Equal at equivalence point)    change to M  

V of OH-:		V = mol = 0.0150 mol =
			     M    0.300 mol
				       L
	Step 2:  CN- + H2O ----> HCN + OH-
		0.0500 L or 50.0 mL of NaOH
need Kb	   0.120 - x              x     x

Kb = Kw/Ka = 1.0 x 10-14/4.0 x 10-10	0.0150 mol/0.125 L = 0.120 M CN-
									   
	Kb = 2.5 x 10-5 = [HCN][OH-]/[CN-] =    x2   = 2.5 x 10-5/0.120 L

	x = 1.7 x 10-3 = [OH-]		pOH = 2.78 		pH = 11.22  

3. (4 pts.)  What is the pH of a buffer solution containing 
	0.800 M HF and 0.500 M NaF?  Ka = 4.0 x 10-4   

       		HF ---->             H+ +    F-			     
At equil.:  0.800-x(ignore x)	     x	   0.500+x(ignore x)

[H+][F-]/[HF] = 4.0 x 10-4 	     x(0.500)/(0.800) = 4.0 x 10-4
	      
	x = 6.4 x 10-4 = [H+]	     pH = 3.19  
								            
4. (3 pts.)  a.	What is the solubility of Mg(OH)2 in water?
	Ksp = 1.2 x 10-11
Mg(OH)2(s)   Mg2+(aq) + 2OH-(aq)  Ksp	= [Mg2+][OH-]2
            x          2x
x(2x)2 	= 1.2 x 10-11
    4x3	= 1.2 x 10-11
x3	= 3.0 x 10-12
      x = 1.4 x 10-4

   (2 pts.)  b. What is the solubility of Mg(OH)2 in 0.10 M NaOH?

		Ksp = 1.2 x 10-11
	Mg(OH)2(s) ----> Mg+2(aq) + 2OH-(aq)	
	                 x         0.10 + 2x(ignore 2x)	
	Ksp = [Mg2+][OH-]2	  x(0.10)2 = 1.2 x 10-11
			 x = 1.2 x 10-9 M 
5. (4 pts.)  Calculate the [H+] of a solution prepared by
mixing 60.0 mL of 2.50 M KOH(SB) and 40.0 mL of 1.75 M H2SO4(SA).


0.0600 L x 2.50 mol/L = 0.150 mol  KOH
0.0400 L x 1.75 mol/L = 0.0700 mol H2SO4



            2KOH    +     H2SO4 ----> 2H2O + K2SO4

 Initial  0.150 mol	 0.0700
*Reaction-0.140 mol	-0.0700
 Final	  0.010 mol      0.0000
 change to M	0.010 mol/0.100 L = 0.10 M KOH or = 0.10 M [OH-]

*0.0700 mol H2SO4 x 2 mol KOH/1 mol H2SO4 = 0.140 mol KOH

	[H+] = Kw/[OH-] = 1.0 x 10-14/0.10 M [OH-] = 1.0 x 10-13 M