Chem 112 Class Guide: pH (Chapter 16, Sections 4-6)
Chapter 16, Sections 4-6 Learning Goals:
Upon completion of Chapter 16, Sections 4-6, you should be able to determine the following:
Calculate pH or pOH given appropriate concentrations
Identify an acid as strong or weak
Calculate the pH of a strong acid
Calculate the pH of a strong base
Calculate the pH of a weak acid
Calculate the percent ionization of a weak acid
Chapter reading Guide: Chapter 16, Sections 4-6
Section 4: THE pH SCALE
Read Chapter 16.4
pH is an arbitrary way to express how much H+ is in a solution.
pH = -log [H+]; pOH = -log[OH-]
pH = 7 – neutral solution
pH < 7 – acidic solution
pH > 7 – basic solution
pH + pOH = 14
Try Practice Exercise 16.6
Try Practice Exercise 16.7
Section 5: STRONG ACIDS AND BASES
Read Chapter 16.5
Strong acids and bases completely dissociate into ions in water.
There are 7 strong acids that you must memorize: HCl, HBr, HI, HNO3, HClO3, HClO4 and
H2SO4
The strong bases are any group 1 or group 2 hydroxides.
Try Practice Exercise 16.8
Try Practice Exercise 16.9
Section 6: WEAK ACIDS (pages 666-673)
Read Chapter 16.6 UP TO PAGE 673 ONLY
A weak acid is any acid that is not a strong acid. They do not completely dissociate, but
instead reach equilibrium with their component ions.
To calculate the pH of weak acids, you must set up an ICE chart as we did in Chapter 15.
Since you do not know how much the initial concentrations will change, there will be “X”s in
the C row of the ICE chart, like in Section 15.6. Refer to that class guide if you are having
trouble remembering how to do this.
HOWEVER, if the K for the acid is less than 1 x 10-2 (which most will be), then you do not
need to use the quadratic equation to solve for X. We can ignore X in the E row of the ICE
chart for the acid ONLY. Instead of: E = (initial concentration – X) you have for your
reactant, you can simplify that to become E = the initial concentration. Once you solve for X,
the value of it will be so small as to be negligible when compared to the initial concentration
of acid. Solving for X is then just simple algebra and taking the negative log of X will yield
the pH of the solution.
Example:
For the reaction HF (aq)  H+ (aq) + F- (aq), Ka = 6.8 x 10-4. Calculate the pH of the solution if
the initial concentration of HF is 0.5 M.
I
C
E
HF (aq) 
0.5
-X
0.5 – X
H+ (aq) +
0
+X
X
F- (aq)
0
+X
X
Now, we solve for X using the K:
 H    F    X  X 
K

 6.8 x104
0.5  X
 HF 
But since K has a value less than 1 x 10-2, we can ignore the –X for the equilibrium value of
HF, and K simplifies to:
 H    F    X  X 
K

 6.8 x104
0.5
 HF 
Solving for X, we get:
X 2  3.4 x104 ; then X  3.4 x104  0.018
So the equilibrium concentrations of each species are:
HF = 0.5 – 0.018 = 0.482 = 0.5 with significant figures! So it was safe to ignore that –X;
H+ = F- = 0.018 M.
So to calculate pH:
pH   log  0.018  1.74
Try Practice Exercise 16.10
Try Practice Exercise 16.11
Try Practice Exercise 16.12
Try Practice Exercise 16.13
Learning Resources
Chapter Learning Goals
Chapter 16, Sections 4-6 Learning Goals
Pre Class Assignment: This assignment must be completed prior to the next class. Check your syllabus
for the exact due date and time.
Complete the pre class assignment
(http://berks.psu.edu/clt/chem112/pH_HW.docx)
Submit a copy to the dropbox located in ANGEL called “Pre Class Assignment Submission: pH”
End of Chapter Problems:
Practice with these problems if you are having difficulty with any of the concepts covered in this
class guide AFTER we have met in class. If you cannot easily complete these problems, seek
help from your instructor, your mentor or the learning center
Chapter 16: 39, 41, 43, 47, 51, 55