# The pH Scale, cont.

Recall

pH = -log[H+]

and

• Range: pH = -1 to pH = 15 (10M -10-15M)
• At midrange [H+] = [OH-] = 10-7M. The solution is said to be "neutral."
• This follows in aqueous solution from Kw = 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 10-7M 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 10-7M OH-, (1.0 x 10-14) / (10-7) = 1.0 x 10-7 and pH = -log(1.0 x 10-7) = 7

Examples:

• What is the pH of a solution of 0.015 M HCl?
Strong acid, so [H+] = 0.015 M
pH = - log [H+]
pH = - log 0.015 = - (- 1.824)
pH = 1.82

Note that the significant figures are correct, 1 is the power of ten, only the figures to the right are significant.

• What is the pH of a solution of 0.067 M NaOH
Strong base, so [OH-] = 0.067 M
Recall that [H+][OH-] = 1.0 x 10-14
Substituting, [H+][0.067] = 1.0 x 10-14
Rearranging, [H+] = (1.0 x 10-14) / 0.067 = 1.493 x 10-13
pH = - log (1.493 x 10-13) = - (- 12.83)
pH = 12.83
Again note the significant figures - 12 corresponds to the power of ten, only the figures to the right are significant.

Note that the "p" has the more general meaning of "-log[]". Thus pOH is -log [OH-], pCa = -log [Ca2+], etc.

## pH of weak acid solutions

Weak acid dissociations involve equilibria. The equilibrium constants have a specific symbol = Ka.

Example: What is the pH of a 0.10 M solution of acetic acid. Ka = 1.8 x 10-5

 HOAc H+ + OAc- Before reaction 0.10 M 0 0 @ Equilibrium 0.10 M- x x x

Ka = [H+][OAc-] / [HOAc]

assume x << 0.1 since Ka =1.8 x 10-5, then [HOAc] = 0.10 M

Substituting, Ka = (x)(x) / 0.10 = 1.8 x 10-5,

x2 = 1.8 x 10-6

x = 1.34 x 10-3M; 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. Ka = 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

Ka = [H+][OAc-] / [HOAc]

Substituting, Ka = [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

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