Acid-Base Calculations 3

When initial concentrations are known

When you know the initial concentration of your acid or base, you can solve the ICE table for your system as you have done for “general” equilibria. Remember that if you know the identity of the acid or base, you also know it’s Ka (or Kb), since you can look it up in tables for acids and bases.

You can also use the same shortcuts and “small-x” approximation as for general equilibrium, following the same rules.
 

Calculating Equilibrium Concentrations in a Weak Acid Solution
Formic acid, HCO2H, is one irritant that causes the body’s reaction to some ant bites and stings.


The pain of some ant bites and stings is caused by formic acid. (credit: John Tann)

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

$$HCO_2H(aq)+H_2O(l)⇌H_3O^+(aq)+HCO_2^-(aq)\qquad K_a=1.8×10^{-4}$$

Solution

The ICE table for this system is

Substituting the equilibrium concentration terms into the Ka expression gives

$$K_a=1.8×10^{-4}=\frac{[H_3O^+][HCO_2^-]}{[HCO_2H]}$$ $$=\frac{(x)(x)}{0.534-x}=1.8×10^{-4}$$



The relatively large initial concentration and small equilibrium constant permits the simplifying assumption that x will be much lesser than 0.534, and so the equation becomes

$$K_a=1.8×10^{-4}=\frac{x^2}{0.534}$$



Solving the equation for x yields

$$x^2=0.534\times (1.8×10^{-4})=9.6×10^{-5}$$ $$x=\sqrt{9.6×10^{-5}}$$ $$x=9.8×10^{-3}\;M$$

To check the assumption that x is small compared to 0.534, its relative magnitude can be estimated:

$$\frac{x}{0.534}=\frac{9.8×10^{-3}}{0.534}=1.8×10^{-2}\;\text{1.8% of 0.534}$$

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

As defined in the ICE table, x is equal to the equilibrium concentration of hydronium ion:

$$x=[H_3O^+]=0.0098\;M$$

Finally, the pH is calculated to be

$$\text{pH}=-log[H_3O^+]=-log(0.0098)=2.01$$

Check Your Learning

Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of a 0.100-M solution of acetic acid, CH3CO2H?

$$CH_3CO_2H(aq)+H_2O(l)⇌H_3O^+(aq)+CH_3CO_2^-(aq)\qquad K_a=1.8×10^{-5}$$
Answer:

percent ionization = 1.3%

Calculating Equilibrium Concentrations in a Weak Base Solution
Find the concentration of hydroxide ion, the pOH, and the pH of a 0.25-M solution of trimethylamine, a weak base:

$$(CH_3)_3N(aq)+H_2O(l)⇌(CH_3)_3NH^+(aq)+OH^-(aq)\qquad K_b=6.3×10^{-5}$$

Solution
The ICE table for this system is


Substituting the equilibrium concentration terms into the Kb expression gives

$$K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}=\frac{(x)(x)}{0.25-x}=6.3×10^{-5}$$



Assuming x << 0.25 and solving for x yields

$$x=4.0×10^{-3}\;M$$



This value is less than 5% of the initial concentration (0.25), so the assumption is justified.

As defined in the ICE table, x is equal to the equilibrium concentration of hydroxide ion:

$$[OH^-]=~0+x=x=4.0×10^{-3}\;M$$

 



The pOH is calculated to be

$$\text{pOH}=-log(4.0×10^{-3})=2.40$$



Using the relation introduced in the previous section of this chapter:

$$\text{pH}=\text{pOH}=pK_w=14.00$$



permits the computation of pH:

$$\text{pH}=14.00-\text{pOH}=14.00-2.40=11.60$$

Check Your Learning

Calculate the hydroxide ion concentration and the percent ionization of a 0.0325-M solution of ammonia, a weak base with a Kb of $1.76×10^{-5}$.

Answer:

$7.56×10^{-4}\;M$, 2.33%