8.7 Acid-base titrations  (Page 9/8)

As seen in the chapter on the stoichiometry of chemical reactions, titrations can be used to quantitatively analyze solutions for their acid or base concentrations. In this section, we will explore the changes in the concentrations of the acidic and basic species present in a solution during the process of a titration.

Titration curve

Previously, when we studied acid-base reactions in solution, we focused only on the point at which the acid and base were stoichiometrically equivalent. No consideration was given to the pH of the solution before, during, or after the neutralization.

Calculating ph for titration solutions: strong acid/strong base

(a) 0.00 mL

(b) 12.50 mL

(c) 25.00 mL

(d) 37.50 mL

Solution

The total initial amount of the hydronium ions is:

Once X mL of the 0.100- M base solution is added, the number of moles of the OH ⁻ ions introduced is:

The total volume becomes: $V=(\text{25.00 mL}+\text{X mL})\left(\frac{\text{1 L}}{\text{1000 mL}}\right)$

The number of moles of H ₃ O ⁺ becomes:

The concentration of H ₃ O ⁺ is:

The preceding calculations work if $\text{n}{({\text{H}}^{\text{+}})}_0-\text{n}{({\text{OH}}^{\text{−}})}_0>0$ and so n(H ⁺ )>0. When $\text{n}{({\text{H}}^{\text{+}})}_0=\text{n}{({\text{OH}}^{\text{−}})}_0,$ the H ₃ O ⁺ ions from the acid and the OH ⁻ ions from the base mutually neutralize. At this point, the only hydronium ions left are those from the autoionization of water, and there are no OH ⁻ particles to neutralize them. Therefore, in this case:

Finally, when $\text{n}{({\text{OH}}^{\text{−}})}_0>\text{n}{({\text{H}}^{\text{+}})}_0,$ there are not enough H ₃ O ⁺ ions to neutralize all the OH ⁻ ions, and instead of $\text{n}({\text{H}}^{\text{+}})=\text{n}{({\text{H}}^{\text{+}})}_0-\text{n}{({\text{OH}}^{\text{−}})}_0,$ we calculate: $\text{n}({\text{OH}}^{\text{−}})=\text{n}{({\text{OH}}^{\text{−}})}_0-\text{n}{({\text{H}}^{\text{+}})}_0$

In this case:

Let us now consider the four specific cases presented in this problem:

(a) X = 0 mL