14.3 Relative strengths of acids and bases  (Page 6/18)

Equilibrium concentrations in a solution of a weak acid

What is the concentration of hydronium ion and the pH in a 0.534- M solution of formic acid?

Solution

1. Determine x and equilibrium concentrations . The equilibrium expression is:
   
   ${\text{HCO}}_2\text{H}(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{HCO}}_2{}^{\text{−}}(aq)$
   
   The concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its change in concentration when setting up the ICE table.
   The table shows initial concentrations (concentrations before the acid ionizes), changes in concentration, and equilibrium concentrations follows (the data given in the problem appear in color):
   
2. Solve for x and the equilibrium concentrations. At equilibrium:
   
   $K_{\text{a}}=1.8\times {10}^{\mathrm{-4}}=\frac{[{\text{H}}_3{\text{O}}^{\text{+}}][{\text{HCO}}_2{}^{\text{−}}]}{[{\text{HCO}}_2\text{H}]}$
   $=\frac{\left(x\right)\left(x\right)}{0.534-x}=1.8\times {10}^{\mathrm{-4}}$

Now solve for x . Because the initial concentration of acid is reasonably large and K _a is very small, we assume that x <<0.534, which permits us to simplify the denominator term as (0.534 − x ) = 0.534. This gives:

Solve for x as follows:

To check the assumption that x is small compared to 0.534, we calculate:

x is less than 5% of the initial concentration; the assumption is valid.

We find the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration as indicated in the last line of the table:

The pH of the solution can be found by taking the negative log of the $[{\text{H}}_3{\text{O}}^{\text{+}}],$ so:

Check your learning

(Hint: Determine $[{\text{CH}}_3{\text{CO}}_2{}^{\text{−}}]$ at equilibrium.) Recall that the percent ionization is the fraction of acetic acid that is ionized $\times$ 100, or $\frac{[{\text{CH}}_3{\text{CO}}_2{}^{\text{−}}]}{{[{\text{CH}}_3{\text{CO}}_2\text{H}]}_{\text{initial}}}\times 100.$

The following example shows that the concentration of products produced by the ionization of a weak base can be determined by the same series of steps used with a weak acid.

Equilibrium concentrations in a solution of a weak base

Find the concentration of hydroxide ion in a 0.25- M solution of trimethylamine, a weak base:

${({\text{CH}}_3)}_3\text{N}(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {({\text{CH}}_3)}_3{\text{NH}}^{\text{+}}(aq)+{\text{OH}}^{\text{−}}(aq)K_{\text{b}}=6.3\times {10}^{\mathrm{-5}}$

Solution

This problem requires that we calculate an equilibrium concentration by determining concentration changes as the ionization of a base goes to equilibrium. The solution is approached in the same way as that for the ionization of formic acid in [link] . The reactants and products will be different and the numbers will be different, but the logic will be the same:

1. Determine x and equilibrium concentrations . The table shows the changes and concentrations:
   
2. Solve for x and the equilibrium concentrations . At equilibrium:
   
   $K_{\text{b}}=\frac{[{({\text{CH}}_3)}_3{\text{NH}}^{\text{+}}]\left[{\text{OH}}^{\text{−}}\right]}{[{({\text{CH}}_3)}_3\text{N}]}=\frac{\left(x\right)\left(x\right)}{0.25-x}=6.3\times {10}^{\mathrm{-5}}$
   
   If we assume that x is small relative to 0.25, then we can replace (0.25 − x ) in the preceding equation with 0.25. Solving the simplified equation gives:
   
   $x=4.0\times {10}^{\mathrm{-3}}$
   
   This change is less than 5% of the initial concentration (0.25), so the assumption is justified.
   Recall that, for this computation, x is equal to the equilibrium concentration of hydroxide ion in the solution (see earlier tabulation):
   
   $\left[{\text{OH}}^{\text{−}}\right]=~0+x=x=4.0\times {10}^{\mathrm{-3}}M$
   
   
   $=4.0\times {10}^{\mathrm{-3}}M$
   
   Then calculate pOH as follows:
   
   $\text{pOH}=\text{−log}\left(4.0\times {10}^{\mathrm{-3}}\right)=2.40$
   
   Using the relation introduced in the previous section of this chapter:
   
   $\text{pH}+\text{pOH}=\text{p}{K}_{\text{w}}=14.00$
   
   permits the computation of pH:
   
   $\text{pH}=14.00-\text{pOH}=14.00-2.40=11.60$
3. Check the work . A check of our arithmetic shows that K _b = 6.3 $\times$ 10 ⁻⁵ .

Check your learning

(a) Show that the calculation in Step 2 of this example gives an x of 4.0 $\times$ 10 ⁻³ and the calculation in Step 3 shows K _b = 6.3 $\times$ 10 ⁻⁵ .

(b) Find the concentration of hydroxide ion in a 0.0325- M solution of ammonia, a weak base with a K _b of 1.76 $\times$ 10 ⁻⁵ . Calculate the percent ionization of ammonia, the fraction ionized $\times$ 100, or $\frac{[{\text{NH}}_4{}^{\text{+}}]}{[{\text{NH}}_3]}\times 100$

Answer:

7.56 $\times$ 10 ⁻⁴ M , 2.33%

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