elmochem Ms. Elmo
Sanderson@Wakefield High School  


1. The student will be able to define equivalent and equivalent mass.
2. The student will be able to find the number of equivalents and the equivalent mass for a given acid or base in an acid-base reaction.
3. The student will be able to describe self-ionization of water.
4. The student will be able to define pH and give the pH of a neutral solution.
5. The student will be able to explain and use the pH scale.
6. The student will be able to calculate pH if [H3O+] or [OH-] is given.
7. The student will be able to calculate [H30+] or [OH-] if pH is given.
8. The student will be able to describe how an acid-base indicator functions.
9. The student will be able to explain the procedures involved in carrying out an acid-base titration.
10. The student will be able to calculate the molarity of a solution from titration data.

Homework:  Due the day of the test.  The day of the test is April 14.

Reviewing concepts: p. 517  1a, 2a, 5a-c, 7, 11

Problems: p. 517-519  1c, 2c, 8c, 9c, 10c, 11c, 12c, 13c, 14c, 15c, 17a, 18, 19

Extra Credit: Worth 10 points on your Chapter 17 test.

Problems: p. 517-519  1b, 2b, 8c, 9b, 10b, 11b, 12b, 17c, 20, 24


17.1 Concentration Units for Acids and Bases

Equivalents: quantities of solutes that have equivalent combining capacity.
HCl  +  KOH  ®  KCl  + H2O
1 mol  1 mol 1 mol 1 mol

½ H2SO4  + KOH  ®  ½ K2SO4  + H2O
½ mol 1 mol   ½ mol      1 mol

1 mol of HCl and ½ mol H2SO4 are chemically equivalent in neutralization reactions with KOH.  That is they both react completely with the same quantity of KOH (1mol).

Equivalent mass acid:  the mass of the acid in grams that will provide one mole of protons ions in a reaction.

Equivalent mass of base:  the mass of the base in grams that will provide one mole of hydroxide ions in a reaction.

Example Problem 1.
Determine the mass of one equivalent of HCl in neutralization reactions.

36.5 g HCl x 1 mol HCl = 36.5 g HCl/ mol H+
1 mol HCl      1 mol H+

Example Problem 2.
Determine the mass of one equivalent of H2SO4 in neutralization reactions.

98.1g H2SO4 x 1 mol H2SO4 = 49.0 H2SO4/ mol H+
1 mol H2SO4   2 mol H+

Works the same way with bases and hydroxide ions!!!!!!

17.2 Aqueous Solutions and the Concept of pH.

Self Ionization of water:  two water molecules interact to produce a hydronium ion (H3O+) and a hydroxide ion by proton transfer.

H2O (l) + H2O (l)  ® H3O+ (aq)  + OH- (aq)

The pH Scale

"pH"  is French for pouvoir hydrogene, meaning "hydrogen power"

pH is defined as the negative of the common (base 10) logarithm of the hydronium ion concentration [H3O+].  The pH is expressed by the following equation:

pH = - log [H3O+]

If pH < 7 the solution is acidic. If pH > 7 the solution is basic.

The pH range generally falls between 0 and 14.

Example Problem 3.
What is the pH of a 0.001 M NaOH solution?

0.001M NaOH = 0.001M [OH-]
pOH = -log(0.001) = 3
pH= 14-pOH
pH= 11

Practice Problem 1.
Determine the pH of a 1 x 10 -3 M HCl solution.

Answer: pH=3

Example Problem 4.
Determine the hydronium ion concentration of an aqueous solution that has a pH of 4.
4= - log x antilog is usually 2nd function log button
antilog(-4) = x
x= 1 x 10 -4M

Practice Problem 2.
The pH of a solution is measured and determined to be 7.52.  What is the hydronium ion concentration?

Answer: 3.02 x 10-8 M

Bases:  Work the same as acid expect they have pOH and you use the [OH-].

Practice Problem 3.
Calculate the pOH of a 3.4 x 10-4M NaOH solution.

Answer: pOH=3.47

We can calculate the pOH from the pH and then determine the [H3O+] or the [OH-].  
pH + pOH = 14

Example Problem 5.

The pH of a solution is determined to be 7.52.  (a) What is the pOH?  (b) What is the [OH-]?
(a) pOH= 14-7.52=6.48 (b) 6.48= -log x
           antilog (-6.48)=x

If more than one hydrogen ion or hydroxide ion you must multiply the concentration by the number of hydrogen or hydroxide ion.

17.3 Acid-Base Titrations.

Indicators:  are weak acid or base dyes whose colors are sensitive to pH or hydronium ion concentration.

In other words, they change color according to the pH of the solution.

Indicators are used in titrations so you know when a certain pH has been reached.

Titration:  the controlled addition and measurement of the amount of a solution of known concentration that is required to react completely with a measured amount of a solution of unknown concentration.

Titration Vocabulary:

Buret: the piece of equipment used to gradually add titrant to the analyte in a titration.

Titrant:  typically the solution of known concentration.  Goes in buret.

Analyte: typically the solution of unknown concentration. Goes in flask.

Equivalence point: the point when there are equivalent quantities of hydronium and hydroxide ions.

End point: the point in a titration where an indicator changes color.

Molarity and Titration.

To find the molarity of an unknown solution using titration:

1. Determine the moles of acid (or base) from the standard solution used during the titration.
2. From a balanced chemical equation, determine the ratio of moles of acid (or base) to base (or acid).
3. Determine the moles of solute of the unknown solution used during the titration.
4. Determine the molarity of the unknown solution.

Example Problem 6.
In a titration, 27.4 mL of a standard solution of Ba(OH)2 is added to a 20.0 mL sample of HCl solution.  The concentration of the standard solution is 0.0154 M.  What is the molarity of the acid solution?

Step 1: Volume of solution used ® moles of base used

0.0154mol Ba(OH)2x 0.0274L x = 4.22 x 10-4mol Ba(OH)2

Step 2: Balanced chemical equation® ratio of moles of acid to moles
  of base.

Ba(OH)2 + 2 HCl ® BaCl2 + 2H2O

1 mol Ba(OH)2 : 2 mol HCl

Step 3: Mole ratio, moles of base used® moles of acid used from unknown solution.

2 mol HClx4.22 x 10-4 molBa(OH)2=8.44 x 10-4 mol HCl
1 mol Ba(OH)2

Step 4: Volume of unknown solution, moles of solute in unknown solution ® molarity of unknown solution

moles of solute in unknown solution x 1000 mL
Volume of unknown solution in mL 1 L

= molarity of unknown solution

8.44X 10-4 moles HCl x 1000 mL = 0.0422 M HCl
         20 mL           1 L

Practice Problem 6.
In a 15.5 mL sample of 0.215 M KOH solution required 21.2 mL of aqueous acetic acid solution in a titration experiment.  Calculate the molarity of the acetic acid solution.

ANSWER: 0.157M HC2H3O2

Practice Problem 7.
By titration, 17.6 mL of aqueous H2SO4 just neutralized 27.4 mL of 0.0165 M LiOH solution.  What was the molarity of the aqueous acid solution?

ANSWER: 0.0128M H2SO4
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Last updated  2008/09/28 10:12:17 PDTHits  2744