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pH Weak Acids and Bases

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WEAK ACIDS

Basic Information

1) Weak acids are less than 100% ionized in solution.
2) Acetic acid (formula = HC2H3O2) is the most common weak acid example used by instructors.
3) Another way to write acetic acid's formula is CH3COOH.
4) A common abbreviation for acetic acid is HAc, where Ac¯ refers to the acetate polyatomic ion.

 

The following equation describes the reaction between acetic acid and water:

Vinegar is a dilute water solution of acetic acid with small amounts of other components. Calculate the pH of bottled vinegar that is 0.667 M HC2H3O2, assuming that none of the other components affect the acidity of the solution.

HC2H3O2(aq)      H+(aq) + C2H3O2-(aq)

We get the value for the acid dissociation constant for this reaction from the table above.

**x in the denominator, is considered too small compared to 0.667, so it is ignored.

x2 = 1.2 x 10-5 x = 3.5 x 10-3

[H+] = 3.5 x 10-3 M H+       pH = -log(3.5 x 10-3) = 2.46


WEAK BASES

A typical pH problem

Calculate the pH and percentage protonation of a 0.20 M aqueous solution of pyridine, C5H5N.

The Kb for C5H5N is 1.8 x 10-9

First, write the proton transfer equilibrium:

\mathrm{H_2O(l) + C_5H_5N(aq) \leftrightarrow C_5H_6N^+ (aq) + OH^- (aq)}

K_b=\mathrm{[C_5H_6N^+][OH^-]\over [C_5H_5N]}

The equilibrium table, with all concentrations in moles per liter, is

  C5H5N C5H6N+ OH-
initial normality .20 0 0
change in normality -x +x +x
equilibrium normality .20 -x x x


Substitute the equilibrium molarities into the basicity constant K_b=\mathrm {1.8 \times 10^{-9}} = {x \times x \over .20-x}
Assume that x << .20. \mathrm {1.8 \times 10^{-9}} \approx {x^2 \over .20}
Solve for x. \mathrm x \approx \sqrt{.20 \times (1.8 \times 10^{-9})} = 1.9 \times 10^{-5}
Check the assumption that x << .20 \mathrm 1.9 \times 10^{-5} \ll .20; so the approximation is valid
Find pOH from pOH = -log [OH-] with [OH-]=x \mathrm pOH \approx -log(1.9 \times 10^{-5}) = 4.7
From pH = pKw - pOH, \mathrm pH \approx 14.00 - 4.7 = 9.3

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