Chem 109  General Chemistry  Spring 2011
Lecture Notes 39: 29 April
The pH Scale, cont.
Recall
pH = log[H^{+}]
and
 Range: pH = 1 to pH = 15 (10M 10^{15}M)
 At midrange [H^{+}] = [OH^{}] = 10^{7}M. The solution is said to be "neutral."
 This follows in aqueous solution from K_{w} = 1.0 x 10^{14} = [H^{+}] [OH^{}], thus if [H^{+}] = [OH^{}], then [H^{+}] = (1.0 x 10^{14})^{1/2}= 1.0 x 10^{7}
 Low pH means acidic:
 For 1M strong acid, pH = 0.0 (log 1 = 0)
 For 0.1M strong acid, pH = 1.0
 For 10^{7}M H^{+}, pH = 7
 High pH means basic:
 For 1M strong base, pH = 14 ([H^{+}] = (1.0 x 10^{14}) / [OH^{}] = (1.0 x 10^{14}) / 1 = 1.0 x 10^{14} and pH = log(1.0 x 10^{14}) = 14.0.
 For 0.1 M OH^{}, (1.0 x 10^{14}) / 0.1 = 1.0 x 10^{13} and pH = log(1.0 x 10^{13}) = 13.0.
 For 10^{7}M OH^{}, (1.0 x 10^{14}) / (10^{7}) = 1.0 x 10^{7} and pH = log(1.0 x 10^{7}) = 7
Examples:
Note that the "p" has the more general meaning of "log[]". Thus pOH is log [OH^{}], pCa = log [Ca^{2+}], etc.
pH of weak acid solutions
Weak acid dissociations involve equilibria. The equilibrium constants have a specific symbol = K_{a}.
Example: What is the pH of a 0.10 M solution of acetic acid. K_{a} = 1.8 x 10^{5}

HOAc 

H^{+} 
+ 
OAc^{} 
Before reaction 
0.10 M 

0 

0 
@ Equilibrium 

0.10 M x


x 

x 
K_{a} = [H^{+}][OAc^{}] / [HOAc]
assume x << 0.1 since K_{a} =1.8 x 10^{5}, then [HOAc] = 0.10 M
Substituting, K_{a} = (x)(x) / 0.10 = 1.8 x 10^{5},
x^{2} = 1.8 x 10^{6}
x = 1.34 x 10^{3}M; assumption OK.
pH =  log (1.34 x 10^{3}) = 2.87
Notice the significant figures. For a log function the number in front of the decimal is the exponent of ten, thus pH = 2.87 is a 2 significant figure number.
Acid Equilibria
Buffer calculations: One of the most frequent calls for calculating acid equilibria is calculations involving buffers. What is a buffer?
 A buffer is a solution which resists changes in pH. Essentially it consists of an acid and its salt (an acid and its conjugate base) in solution together. Thus the solution has a proton donor and a proton acceptor, so pH is stabilized.
 A buffer is simply an acid equilibrium system with significant amounts of both the acid and its conjugate base.
With this in mind let's do some examples.
Example: Calculate the pH of a buffer made up by dissolving 0.0125 moles acetic acid (HOAc) and 0.0250 moles of sodium acetate (NaOAc) in enough water to make 1.000 L of solution. K_{a} = 1.8 x 10^{5}

HOAc 

H^{+} 
+ 
OAc^{} 
Before reaction 
0.0125 moles/L 

0 

0.0250 moles/L 
@ Equilibrium 


(0.0125  x) M

assume x is small,

= 0.0125


x 



(0.0250 + x) M

assume x is small,

= 0.0250


K_{a} = [H^{+}][OAc^{}] / [HOAc]
Substituting, K_{a} = [H^{+}](0.0250) / (0.0125) = 1.8 x 10^{5}
Rearranging, [H^{+}] = (1.8 x 10^{5})(0.0125) / (0.0250) = 0.90 x 10^{5}
x is within experimental error (0.000009 < ±0.0001), so assumption OK
pH = 5.046
© R A Paselk
Last modified 27 April 2011