8.7 Acid-base titrations  (Page 3/8)

Although the initial volume and molarity of the acids are the same, there are important differences between the two titration curves. The titration curve for the weak acid begins at a higher value (less acidic) and maintains higher pH values up to the equivalence point. This is because acetic acid is a weak acid, which is only partially ionized. The pH at the equivalence point is also higher (8.72 rather than 7.00) due to the hydrolysis of acetate, a weak base that raises the pH:

After the equivalence point, the two curves are identical because the pH is dependent on the excess of hydroxide ion in both cases.

Titration of a weak acid with a strong base

(a) What is the initial pH before any amount of the NaOH solution has been added? K _a = 1.8 $\times$ 10 ⁻⁵ for CH ₃ CO ₂ H.

(b) Find the pH after 25.00 mL of the NaOH solution have been added.

(c) Find the pH after 12.50 mL of the NaOH solution has been added.

(d) Find the pH after 37.50 mL of the NaOH solution has been added.

Solution

$K_{\text{a}}=\frac{[{\text{H}}_3{\text{O}}^{\text{+}}][{\text{CH}}_3{\text{CO}}_2{}^{\text{−}}]}{[{\text{CH}}_3{\text{CO}}_2\text{H}]}\approx \frac{{[{\text{H}}_3{\text{O}}^{\text{+}}]}^{\text{2}}}{{[{\text{CH}}_3{\text{CO}}_2\text{H}]}_0},$ and $[{\text{H}}_3{\text{O}}^{\text{+}}]=\sqrt{K_a\times [{\text{CH}}_3{\text{CO}}_2\text{H}]}=\sqrt{1.8\times {10}^{\mathrm{-5}}\times 0.100}=1.3\times {10}^{\mathrm{-3}}$

(b) After 25.00 mL of NaOH are added, the number of moles of NaOH and CH ₃ CO ₂ H are equal because the amounts of the solutions and their concentrations are the same. All of the CH ₃ CO ₂ H has been converted to ${\text{CH}}_3{\text{CO}}_2{}^{\text{−}}.$ The concentration of the ${\text{CH}}_3{\text{CO}}_2{}^{\text{−}}$ ion is:

The equilibrium that must be focused on now is the basicity equilibrium for ${\text{CH}}_3{\text{CO}}_2{}^{\text{−}}:$

so we must determine K _b for the base by using the ion product constant for water:

Since K _w = [H ⁺ ][OH ⁻ ]:

Let us denote the concentration of each of the products of this reaction, CH ₃ CO ₂ H and OH ⁻ , as x . Using the assumption that x is small compared to 0.0500 M , $K_{\text{b}}=\frac{x^{\text{2}}}{0.0500M},$ and then:

Note that the pH at the equivalence point of this titration is significantly greater than 7.

(c) In (a), 25.00 mL of the NaOH solution was added, and so practically all the CH ₃ CO ₂ H was converted into ${\text{CH}}_3{\text{CO}}_2{}^{\text{−}}.$ In this case, only 12.50 mL of the base solution has been introduced, and so only half of all the CH ₃ CO ₂ H is converted into ${\text{CH}}_3{\text{CO}}_2{}^{\text{−}}.$ The total initial number of moles of CH ₃ CO ₂ H is 0.02500L $\times$ 0.100 M = 0.00250 mol, and so after adding the NaOH, the numbers of moles of CH ₃ CO ₂ H and ${\text{CH}}_3{\text{CO}}_2{}^{\text{−}}$ are both approximately equal to $\frac{\text{0.00250 mol}}{2}=\text{0.00125 mol},$ and their concentrations are the same.

Since the amount of the added base is smaller than the original amount of the acid, the equivalence point has not been reached, the solution remains a buffer, and we can use the Henderson-Hasselbalch equation: