Triprotic Acids and Bases Calculations: Videos & Practice Problems

Understanding the pH of weak triprotic acids involves focusing on their acidic and basic forms. Triprotic acids can donate three protons and have four forms: the fully protonated acidic form, two intermediate forms, and the deprotonated basic form. To calculate the pH of the acidic form, we use the acid ionization constant (Ka₁). Conversely, for the basic form, the base ionization constant (Kb₁) is applied. Remember that while triprotic acids have multiple stages of ionization, the primary concern for pH calculations is the initial acidic and basic forms, which are critical for predicting the acid's behavior in solution.

pH of Weak Triprotic Acid Species

When calculating the pH of weak triprotic acids, it is essential to focus on the acidic and basic forms of the species. A triprotic acid can donate three protons (H⁺ ions), resulting in four distinct forms: the fully protonated acidic form, the first intermediate (having lost one proton), the second intermediate (having lost two protons), and the fully deprotonated basic form.

For pH calculations, we primarily consider the acidic form, which retains all three protons, and the basic form, which has lost all protons. To determine the pH of the acidic form, we use the first acid dissociation constant, K_a1. This constant reflects the strength of the acid in its first dissociation step. Conversely, to calculate the pH of the basic form, we utilize the first base dissociation constant, K_B1, which indicates the strength of the base formed after all protons have been lost.

In summary, while there are four forms of a triprotic species, the calculations for pH focus on the acidic and basic forms, utilizing K_a1 for the acidic form and K_B1 for the basic form to accurately assess their respective pH levels.

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?

To calculate the pH of a weak triprotic acid, focus on the fully protonated acidic form. Use the first acid ionization constant (Ka1) to determine the concentration of hydrogen ions (H⁺). The equation for the ionization of the acidic form is:

$H_{3}A\leftrightarrow H_{2}A+H^{+}$

Using the expression for Ka1:

$K_{\mathrm{a1}}=\frac{{\mathrm{[H}}_{2}A]{\mathrm{[H}}^{+}}{{\mathrm{[H}}_3}$

Solve for [H⁺] and then calculate the pH using:

$\mathrm{pH}=-\log \left({\mathrm{[H}}^{+}\right)$

What are the four forms of a triprotic acid?

A triprotic acid has four forms based on the number of protons it has lost:

1. Fully protonated acidic form: This is the form where the acid has all three protons (H₃A).
2. First intermediate form: This form has lost one proton (H₂A^-).
3. Second intermediate form: This form has lost two protons (HA^2-).
4. Fully deprotonated basic form: This is the form where the acid has lost all three protons (A^3-).

These forms are crucial for understanding the acid's behavior in solution and for pH calculations.

Why is Ka1 used to calculate the pH of the acidic form of a triprotic acid?

Ka1 is used to calculate the pH of the acidic form of a triprotic acid because it represents the first ionization step, where the acid donates its first proton. This step has the highest ionization constant, indicating it is the most significant in determining the concentration of hydrogen ions (H⁺) in solution. The equation for the first ionization is:

$H_{3}A\leftrightarrow H_{2}A+H^{+}$

Using Ka1 ensures that the pH calculation accurately reflects the initial dissociation of the triprotic acid.

How do you calculate the pH of the basic form of a weak triprotic acid?

To calculate the pH of the basic form of a weak triprotic acid, use the first base ionization constant (Kb1). The basic form is the fully deprotonated form (A^3-). The equation for the ionization of the basic form is:

$A_3+H^{2}O\leftrightarrow {\mathrm{HA}}_2+{\mathrm{OH}}^{-}$

Using the expression for Kb1:

$K_{\mathrm{b1}}=\frac{{\mathrm{[HA}}_2]{\mathrm{[OH}}^{-}}{{\mathrm{[A}}_3}$

Solve for [OH^-] and then calculate the pOH. Finally, convert pOH to pH using:

$\mathrm{pH}=14-\mathrm{pOH}$

What is the significance of intermediate forms in triprotic acid calculations?

Intermediate forms in triprotic acid calculations represent the stages where the acid has lost one or two protons. These forms are:

1. First intermediate form (H₂A^-)
2. Second intermediate form (HA^2-)

While the primary focus for pH calculations is on the fully protonated acidic form and the fully deprotonated basic form, the intermediate forms are important for understanding the complete ionization process and the acid's behavior in solution. They help in predicting the acid's buffering capacity and the pH range over which the acid can effectively donate or accept protons.