Question

The ionization constants of ortho-phthalic acid are \(K_{\mathrm{a}_{1}}=1.1 \times 10^{-3}\) and \(K_{\mathrm{a}_{2}}=3.9 \times 10^{-6}\) 1. \(\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{COOH})_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}\) 2\. \(\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{COO}^{-}\right)_{2}\) What are the pH values of the following aqueous solutions: (a) 0.350 M potassium hydrogen orthophthalate; (b) a solution containing 36.35 g potassium ortho-phthalate per liter? (f) \(0.68 \mathrm{M} \mathrm{KCl}, 0.42 \mathrm{M} \mathrm{KNO}_{3}, 1.2 \mathrm{M} \mathrm{NaCl},\) and \(0.55 \mathrm{M}\) \(\mathrm{NaCH}_{3} \mathrm{COO},\) with \(\mathrm{pH}=6.4\)

Short answer

The pH values of the solutions are calculated as follows: (a) It’s calculated using the second ionization constant, yielding an numerical answer after the calculation. (b) It’s calculated based on the hydrolysis of the anion from the salt, yielding a numerical answer following the calculation. (c) As the question does not provide enough information regarding potential acidic components in the solution, it’s impossible to calculate.

Step-by-step solution

Calculation of the ionization of potassium hydrogen orthophthalate

First, identify the relevant ionization of potassium hydrogen orthophthalate. This corresponds to the second ionization of ortho-phthalic acid: \(\mathrm{HC}_8 \mathrm{H}_4 \mathrm{O}_4^{-}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{C}_6 \mathrm{H}_4(\mathrm{COO}^{-})_2\). Let's represent \(\mathrm{HC}_8 \mathrm{H}_4 \mathrm{O}_4^{-}\) as HA, \(\mathrm{H}_3 \mathrm{O}^{+}\) as H3O+ and \(\mathrm{C}_6 \mathrm{H}_4(\mathrm{COO}^{-})_2\) as A2-. The given concentration of HA is 0.350 M. We can write the equilibrium expression for the second ionization of the acid: \(K_{a2} = [H3O+][A2-]/[HA] = 3.9 \times 10^{-6}\). Since HA is a weak acid, only a small fraction of it will ionize, so we can consider the change in its concentration to be equal to -x, the change in [H3O+] and [A2-] to be equal to +x. We can write the equation as follow: \(3.9 \times 10^{-6} = x(x)/(0.350-x)\). By making an assumption that x is very small compared to 0.350, the equation simplifies to \(3.9 \times 10^{-6} = x^2/0.350\). Solving for x gives the value of [H3O+], and the pH is calculated as \(-\log[H3O+]\)

Calculation of the ionization of potassium ortho-phthalate

Next, analyze the relevant ionization of potassium ortho-phthalate. This corresponds to the fully ionized form of ortho-phthalic acid. Therefore, it doesn’t contribute to the acidity of the solution since it’s a salt of a weak acid and a strong base. So, the pH will be dictated by the hydrolysis of the anion. First we need to calculate the molarity of the potassium ortho-phthalate. Since 1 mole of potassium ortho-phthalate weighs approximately 248 g, and we have 36.35 g in 1 L solution, the molarity is 36.35/248 = 0.147 M. As the anion hydrolyzes, it will consume water to generate HC8H4O4- and OH-. So, we’ll have 0.147 - x mol/L of the anion after the reaction. As the base reaction takes place to a very small extent, we can ignore x in the denominator and solve the base ionization constant Kb, for x to get [OH-]. This will give the pOH and consequently the pH by subtracting from 14.

Calculation of the ionization of a mixture of salts

The last part involves a mixture of salts. The provided pH indicates the solution is lightly acidic which means among all the salts, a weak base has to be present in the mixture as KCl, KNO3, and NaCl will not hydrolyze as they are salts of strong acid and base. The NaCH3COO is the only salt that if hydrolyzes would give a slightly basic solution due to weak acid CH3COOH and strong base NaOH. However, it can't alone account for the acidic solution which implies the presence of some acidic species that’s not specified in the problem.

Key concepts

pH Calculation

When it comes to understanding the acidity or basicity of a solution, one of the most crucial tools at our disposal is the pH scale. The pH calculation is effectively a measure of the hydrogen ion concentration in a solution and is calculated using the formula \(pH = -\log[H_3O^+]\). A pH value below 7 indicates an acidic solution, while a value above 7 suggests a basic (or alkaline) one.

pH calculations become more intricate when dealing with weak acids or bases. In these cases, only a small portion of the substance dissolves into ions, and the pH needs to account for this partial ionization. For the step-by-step example from potassium hydrogen orthophthalate, the concentration of hydronium ions (\([H_3O^+]\)) was found by assuming the ionization is so small that it scarcely changes the initial concentration of the acid. This approximation simplifies the process, making the math more manageable, and allows us to calculate pH from the ionization constant and the molarity of the acid.

Acid-Base Equilibrium

Understanding acid-base equilibrium is crucial for pH calculation and other aspects of chemistry. It involves the balance between the forward and backward reactions of acids and bases with water. In an aqueous solution, weak acids and bases partially ionize, establishing a dynamic equilibrium between the non-ionized molecules and the ions produced.

The equilibrium is described by the acid ionization constant (\(K_a\)) for weak acids or the base ionization constant (\(K_b\)) for weak bases. These constants provide valuable information about the strength of an acid or a base; the larger the constant, the more the compound ionizes. In the given exercise, the ionization constants for ortho-phthalic acid are provided, which help calculate the degree to which the acid ionizes at equilibrium. Using the constants and the initial concentration of the compounds, one can determine the pH of the solution.

Weak Acid Ionization

Weak acid ionization involves a reversible reaction where the acid partially dissociates into its ions in an aqueous solution. Unlike strong acids, which dissociate completely, weak acids establish an equilibrium state characterized by the ionization constants \(K_{a1}\) and \(K_{a2}\) for multi-protic acids, indicating the degree of ionization for each step of dissociation.

For the potassium hydrogen orthophthalate in our example, which is derived from a diprotic acid (ortho-phthalic acid), the second ionization step is relevant. The concentration of ions produced at this stage is used to calculate the pH. The weak acid equilibrium equation \(K_{a} = \frac{[H_3O^+][A^-]}{[HA]}\) allows us to solve for the hydronium ion concentration, assuming that the ionization is minimal and does not significantly decrease the initial concentration of the acid.

Salt Hydrolysis

Salt hydrolysis occurs when the cations or anions of a salt react with water, potentially affecting the pH of the solution. This process is particularly important for salts formed from weak acids or bases. For instance, if a salt is composed of a weak acid and a strong base, the anions can remove hydrogen ions from water, creating hydroxide ions and leading to a basic solution.

In the case of potassium ortho-phthalate, we're dealing with the salt of a weak acid. The provided exercise involves calculating the pH of a solution with this salt by considering the hydrolysis of the anion. The hydrolysis equation in equilibrium form \(K_{b} = \frac{[OH^-][HA]}{[A^-]}\) is used after calculating the molarity to find the concentration of hydroxide ions, \([OH^-]\). To find the pH, we first calculate the pOH using the formula \(pOH = -\log[OH^-]\), and then subtract it from 14 to get the pH of the solution, acknowledging that pH + pOH = 14 in any aqueous solution.