Chapter 11 Notes Dr. Floyd Beckford

ACID-BASE TITRATIONS
Standardization is the process by which the concentration of a solution. To carry out standardizations, a solution of a primary standard is needed. The standardized solution is called a secondary standard.

There are five major chemical properties of a primary standard:
1. It must not react with or absorb the components of the atmosphere such as water vapor, CO₂ and O₂.
2. It must have a high percentage purity.
3. It must react according to one invariable reaction.
4. It must be soluble in the solvent of interest.
5. It must have a relatively high formula weight (to minimize errors in weighing).

For standardizing acids, the popular primary standard is sodium carbonate, Na₂CO₃.

	H2SO4 + Na2CO3 ---> Na2SO4 + CO2 + H2O
	1 mole  1 mole
	2HCl + Na2CO3 ---> 2NaCl + CO2 + H2O
	2 mole  1 mole

Example 1: Calculate the molarity of a solution of nitric acid if 35.72 mL of the solution neutralizes 0.364 g of Na₂CO₃.

	Mole ratio acid to base = 2:1

	No. of moles Na2CO3 = Mass of Na2CO3/Formula weight of Na2CO3
			= 0.364 g/ 106.0 g/mol
			= 3.43 x 10-3 mol

	From the mole ratio, no. of moles HNO3 = 2 x no. of moles of Na2CO3
					      = 2 x 3.43 x 10-3 mol
					      = 6.87 x 10-3 mol

	Since 35.72 mL of acid was used, 
		then it is this volume that contains 6.87 x 10-3 mol
	
	Molarity = no. of moles/volume
		   = 6.87 x 10-3 mol/0.03572 L
		   = 0.192 M

	NaOH + KHP ---> NaKP + H2O
	1 mole  1 mole

Example 2: Calculate the molarity of a solution of KOH if 20.58 mL of the solution reacts with 0.6128 g of KHP.

	Mole ratio acid to base = 1:1
	No. of moles Na2CO3 = Mass of KHP/Formula weight of KHP
			= 0.6128 g/204.2 g/mol
			= 3.000 x 10-3 mol
	
	From the mole ratio, no. of moles KOH = no. of moles of KHP
					      = 3.000 x 10-3 mol

	Since 20.58 mL of acid was used, 
	then it is this volume that contains 6.87 x 10-3 mol
	
	Molarity = no. of moles/volume
		   = 3.000 x 10-3 mol/0.02058 L
		   = 0.1458 M

Normality = (number of equivalents of solute)/liter of solution = no. eq/L , or
Normality = no. of milliequivalents/liter = no. of meq/mL

One equivalent of an acid is defined as the mass of the acid (expressed in grams) that will provide 1 mole of H⁺ ions or that will react with 1 mole of OH^- ions.

	HCl          --->         H+         +        Cl-
      1 mol (36.46 g)	      1 mol (1.01 g)

For a diprotic acid, one formula unit produces 2 moles of H⁺ ions.

The arguments above holds true for bases as well.

The equivalent weight of an acid or base is obtained by dividing its formula weight in grams by the number of protons or hydroxides in one formula unit of the acid or base respectively.

The basic premise behind the normality concentration scale is that one equivalent of an acid reacts with one equivalent of a base. This is not always the case for, say, the molarity scale where the number of moles of acid and base that reacts is not necessarily the same.

I. CHANGE-IN-OXIDATION-NUMBER (CON) METHOD
Guidelines
1. Write as much of the overall equation as possible
2. Assign oxidation numbers to all elements on both sides of the equation
3. Insert coefficients into the equation to make the total increase and decrease in oxidation numbers equal.
4. Balance the other atoms by inspection.

Example 3: Aluminum reacts with hydrochloric acid to form aqueous aluminum chloride and gaseous hydrogen. Balance the formula unit equation and identify the oxidizing and reducing agents.

	HCl(aq) + Al(s) ---> AlCl3(aq) + H2(g) 

	Increase	Al = 0  ---> Al = +3	Change = +3
	Decrease	H = +1 --->  H = 0	Change = -1

	Equalize the changes:- for Al; 1(+3) = 3 : for H; 3(-1) = 3
	Because there are 2 H in H2, 
		each change must be multiplied by 2.
	So total change = +6 for Al and -6 for H.
	We need 2 Al and 6 H on wazzu each side of the equation
	
	6HCl(aq) + 2Al(s) ---> 2AlCl3(aq) + 3H2(g)

	Increase	Br = -1  --->  Br = 0		Change = +1
	Decrease	Mn = +7  --->  Mn = +2		Change = -5

	Equalize the changes:- for Br; 5(+1) = 5 : for Mn; 1(-5) = -5
	Because there are 2 Br in Br2, 
		each change must be multiplied by 2.
	So total change = +10 for Br and -5 for Mn.
	We need 2 Br and 2 Mn on each side of the equation

	2MnO4-(aq) + H+(aq) + 10Br-(aq) ---> 2Mn2+(aq) + 5Br2(l)+ H2O(l)	

	Balance the O and H

	2MnO4-(aq) + 16H+(aq) + 10Br-(aq) ---> 2Mn2+(aq) + 5Br2(l)+ 8H2O(l)	

Example 5: Balance the following ionic equation.

	Increase	Al = 0  ---> Al = +3	Change = +3
	Decrease	N = +5 --->  N = -3	Change = -8
	
	Equalize the changes:- for Al; 8(+3) = +24 : for N; 3(-8) = -24
	We need 8 Al and 3 N on each side of the equation

	8Al(s) + 3NO3-(aq) + OH-(aq) + H2O ---> 8[Al(OH)4(aq) + 3NH3(g)

I. THE HALF-REACTION METHOD
In this method the redox reaction is split into its reduction and oxidation parts (called half-reactions). The half reactions are then added and the common terms are eliminated. As in the CON method, H+ (acid), OH- (base) or H2O can be added in order to mass balance the equation.

Example 6: Balance the following net ionic equation for the reaction

	(1) Oxidation half-reaction: C2O42- ---> 2CO2 + 2e-					
	(2) Reduction half-reaction: MnO4- + 5e- + 8H+ ---> Mn2+ + 4H2O
	
	The number of reduction electrons = number of oxidation electrons
	Multiply (1) by 5 and (2) by 2
	i.e.	5 C2O42- ---> 10 CO2 + 10 e-	
		 2 MnO4- + 10 e- + 16 H+ ---> 2 Mn2+ + 8 H2O
		
	Adding to two half-reactions give the balanced redox equation
	
	2 MnO4- + 5 C2O42- + 16 H+ ---> 2 Mn2+ + 10 CO2 + 8 H2O

	(1) Oxidation half-reaction: Fe2+ ---> Fe3+ + e-						
	(2) Reduction half-reaction: MnO4- + 5e- + 8H+ ---> Mn2+ + 4H2OO			
	(Notice we have added 8 H+ and 4 H2O to mass balance this equation)
	The number of reduction electrons = number of oxidation electrons
	Multiply (1) by 5 and let (2) remain the same
	i.e. 	5Fe2+ ---> 5Fe3+ + e-
		MnO4- + 5e- + 8H+ ---> Mn2+ + 4H2O

	Adding to two half-reactions give the balanced redox equation
	
	MnO4- + 5Fe2+ + 8H+ ---> Mn2+ + 5Fe3+ + 4H2O

	(1) Oxidation half-reaction: SO2 + 2H2O ---> SO42- + 4H+ + 2e-
	(2) Reduction half-reaction: Br2 + 2e- ---> 2Br- 						
	
	Adding to two half-reactions give the balanced redox equation
	
	Br2(l) + SO2(g) + 2H2O(l) ---> 2Br-(aq) + SO42-(aq) + 4H+(aq)

	(1) Oxidation half-reaction: Zn ---> Zn2+ + 2e-
	(2) Reduction half-reaction: NO3- + 5e- ---> N2 
	(3) (Balancing (2) : 2NO3- + 10e- + 12H+ ---> N2 + 6H2O))

	The number of reduction electrons = number of oxidation electrons
	Multiply (1) by 5 and let (3) remain the same
	i.e.	5Zn ---> 5Zn2+ + 10e-
		2NO3- + 10e- + 12H+ ---> N2 + 6H2O
	
	Adding the two half-reactions
	
	5Zn(s) + 2NO3-(aq) + 12H+(aq) ---> 5Zn2+(aq) + N2(g) + 6H2O

Example 10: What volume of 0.200 M KMnO₄ would be required to oxidize 50.0 mL of 0.100 M KI in acidic solution? Products include Mn²⁺ and I₂.

	Write as much of the equation as possible.
		MnO4- + I- ---> Mn2+ + I2
	
	(1) Oxidation half-reaction: 2I- ® I2 + 2e-
	(2) Reduction half-reaction: MnO4- + 5e- + 8H+ ---> Mn2+ + 4H2O
	
	(Notice we multiplied (1) by 2 and have added 8 H+ and 4 H2O to (2) in order mass balance these equations).
	
	The number of reduction electrons = number of oxidation electrons
	Multiply (1) by 5 and (2) by 2
	i.e. 	10I- ---> 5I2 + 10e-
		2MnO4- + 10e- + 16H+ ---> 2Mn2+ + 8H2O
	Adding to two half-reactions give the balanced redox equation
	
	2MnO4- + 10I- + 16H+ ---> 2Mn2+ + 5I2 + 8H2O
	
	No of moles I- = molarity x volume
		= 0.100 mol/L x 0.050 L = 5 x 10-3 moles
	Mole ratio of MnO4- to I- is 1:5
	So, no of moles MnO4- = 5 x 10-3 / 5 = 1 x 10-3 moles
	Volume MnO4- = no of moles/ molarity 
		  = 1 x 10-3 moles/ 0.200 mol/L = 0.005 L = 5 mL