Triprotic Acids and Bases Calculations: Videos & Practice Problems

Triprotic Acid pH Example

To determine the pH of a 0.225 molar phosphoric acid solution, we start by recognizing that phosphoric acid is a triprotic acid, meaning it can donate three protons (H⁺) in a stepwise manner. The first step involves setting up an ICE (Initial, Change, Equilibrium) chart to analyze the dissociation of the acid in water, following the Brønsted-Lowry definition of acids and bases.

In the first dissociation step, phosphoric acid (H₃PO₄) donates a proton to water (H₂O), forming dihydrogen phosphate (H₂PO₄^-) and hydronium ions (H₃O⁺). The initial concentration of H₃PO₄ is 0.225 M, while the initial concentrations of the products are zero. The change in concentration for the reactants is represented as -x, while the products increase by +x. Thus, the equilibrium concentrations can be expressed as:

H₃PO₄ (initial): 0.225 M, change: -x, equilibrium: 0.225 - x

H₂PO₄^- (initial): 0, change: +x, equilibrium: x

H₃O⁺ (initial): 0, change: +x, equilibrium: x

Next, we set up the equilibrium constant expression for the first dissociation constant (K_a1), which is given as 7.5 x 10^-3. The expression is:

K_a1 = \(\frac{[H_2PO_4^-][H_3O^+]}{[H_3PO_4]}\) = \(\frac{x^2}{0.225 - x}\)

To simplify the calculation, we can apply the 500 approximation rule. This rule states that if the initial concentration divided by K_a1 is greater than 500, we can ignore the -x in the denominator. However, in this case, the ratio is approximately 30, which is less than 500, indicating that we must keep the -x term and solve the equation using the quadratic formula.

Rearranging the equation leads to:

x² + 7.5 x 10^-3x - 0.0016875 = 0

Using the quadratic formula, where a = 1, b = 7.5 x 10^-3, and c = -0.0016875, we find:

x = \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Calculating this gives us two potential solutions for x: 0.0375 and -0.045. Since concentrations cannot be negative, we take x = 0.0375 M, which represents the concentration of H₃O⁺.

Finally, we can calculate the pH using the formula:

pH = -log[H₃O⁺] = -log(0.0375) ≈ 1.43

Thus, the pH of the 0.225 molar phosphoric acid solution is approximately 1.43.

Determine the pH of 0.250 M sodium phosphate, Na₃PO₄. Phosphoric acid, H₃PO₄, contains K_a1 = 7.5 × 10⁻³, K_a2 = 6.2 × 10⁻⁸ and K_a3 = 4.2 × 10⁻¹³.

• A. 12.82
• B. 7.21
• C. 1.18
• D. 12.38

Answer: (A)

Determine the pOH of 0.450 M citric acid, H₃C₆H₅O₇. It possesses K_a1 = 7.4 × 10⁻⁴, K_a2 = 1.7 × 10⁻⁵ and K_a3 = 4.0 × 10⁻⁷.

• A. 7.602
• B. 13.653
• C. 9.230
• D. 12.260

Answer: (D)

Here's what students ask on this topic:

How do you calculate the pH of a weak triprotic acid in its acidic form?

To calculate the pH of a weak triprotic acid in its acidic form, focus on the fully protonated species that contains all three acidic protons. The key step is to use the first acid dissociation constant, $K_{\mathrm{a1}}$, because the initial ionization dominates the pH. Set up the equilibrium expression for the dissociation of the first proton:

$\mathrm{HA}\to H++A-$

Use the expression $K_{\mathrm{a1}}=\frac{\mathrm{[H<mo>+</mo>][A<mo>-</mo>]}}{}\mathrm{[HA]}$ and solve for the concentration of $H+$. Then calculate pH as $\mathrm{pH\; =\; -log[H<mo>+</mo>]}$. This approach assumes the first dissociation is the most significant contributor to acidity, which is valid for weak triprotic acids. Intermediate forms and further dissociations have minimal effect on the initial pH calculation.

What is the role of the base ionization constant (Kb1) in calculating the pH of a weak triprotic acid?

The base ionization constant, $K_{\mathrm{b1}}$, is used to calculate the pH of the basic form of a weak triprotic acid, which is the fully deprotonated species. When the triprotic acid loses all three protons, it forms a base that can accept protons from water, producing hydroxide ions. The equilibrium expression for this base ionization is:

$B+\mathrm{H<mo>2</mo>O}\to \mathrm{HB}++\mathrm{OH}-$

Using $K_{\mathrm{b1}}$, you can find the hydroxide ion concentration, then calculate pOH and finally pH by $\mathrm{pH\; =\; 14\; -\; pOH}$. This calculation is important when the solution contains the basic form of the triprotic acid, such as in titration or equilibrium scenarios where the fully deprotonated species predominates.

Why are only the acidic form and the basic form considered in pH calculations of weak triprotic acids?

In weak triprotic acid pH calculations, the acidic form (fully protonated) and the basic form (fully deprotonated) are primarily considered because they represent the species with the most significant impact on the solution's pH. The intermediate forms, which have lost one or two protons, exist in smaller concentrations and have less influence on the overall hydrogen ion concentration. The first dissociation constant, $K_{\mathrm{a1}}$, governs the acidic form's pH, while the first base ionization constant, $K_{\mathrm{b1}}$, governs the basic form's pH. This simplification allows for more straightforward calculations without significant loss of accuracy, especially in dilute solutions where the intermediate species are less prevalent.

How do the four forms of a triprotic acid differ in terms of protonation?

A triprotic acid has four distinct forms based on the number of protons it has lost. The fully protonated acidic form contains all three acidic protons. The first intermediate form has lost one proton, the second intermediate form has lost two protons, and the basic form is fully deprotonated, having lost all three protons. These forms can be represented as:

$\mathrm{H₃A}(acidic\; form)\; \to \; H$₂A^- (1st intermediate) → HA^2- (2nd intermediate) → A^3- (basic form)

Each form has different acid-base properties, but for pH calculations, the acidic and basic forms are most relevant because they dominate the solution's behavior in terms of proton donation and acceptance.

What is the significance of Ka1 in the behavior of weak triprotic acids?

The first acid dissociation constant, $K_{\mathrm{a1}}$, is crucial in understanding the behavior of weak triprotic acids because it represents the equilibrium constant for the loss of the first proton. This step usually has the largest effect on the pH of the solution since the initial proton dissociation is the most favorable and contributes the most to the hydrogen ion concentration. Subsequent dissociations (Ka2 and Ka3) are typically weaker and less influential on pH. Therefore, $K_{\mathrm{a1}}$ is used to calculate the pH of the acidic form, simplifying the analysis of triprotic acid solutions.