Question

A sample of \(0.1276 \mathrm{~g}\) of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.0633 \mathrm{M} \mathrm{NaOH}\) solution. The volume of base required to reach the equivalence point was \(18.4 \mathrm{~mL}\). (a) Calculate the molar mass of the acid. (b) After \(10.0 \mathrm{~mL}\) of base had been added to the titration, the \(\mathrm{pH}\) was determined to be 5.87 . What is the \(K_{\mathrm{a}}\) of the unknown acid?

Short answer

a) The molar mass of the acid is \(110 \mathrm{g/mol}\), b) The acid dissociation constant of the unknown acid is approximately \( 1.07 \times 10^{-5}\).

Step-by-step solution

Calculate moles of the base

First we need to determine the moles of \(\mathrm{NaOH}\) base that's used in the reaction. We do this by multiplying the volume of the solution by its molarity \(0.0633 \mathrm{M}\). We also need to convert the volume from milliliters to liters by dividing by 1000:\[0.0633 \mathrm{M} \times \dfrac{18.4 \mathrm{ml}}{1000 \mathrm{ml/L}} = 0.00116 \mathrm{moles}\]

Determine molar mass of the acid

Next, we will use the molarity of the acid to determine the molar mass. We do this by dividing mass by number of moles. So the molar mass of the unknown acid would be: \[\frac{0.1276 \mathrm{g}}{0.00116 \mathrm{mol}}= 110 \mathrm{g/mol}\]

Calculate the acid dissociation constant (\(K_a\))

When the \(\mathrm{pH}\) was determined, it was non-equivalent because only \(10.0 \mathrm{mL}\) of the base was added. So at that point, the acid has not fully dissociated. This point in the reaction can be described by the formula: \[K_a=(x)(x)/(0.00516-x)\] Where x is the concentration of \(OH^-\) ions. We first calculate the \(OH^-\) concentration using the \(pH\) and \(pOH\) relationship: \[pOH= 14 - pH = 14 - 5.87 = 8.13\] Then, we calculate the concentration by using the formula: \[ [OH^-]= 10^{-pOH}= 10^{-8.13}= 7.4 \times 10^{-9}\] Therefore, x is \(7.4 \times 10^{-9}\). After substituting x into equation we can solve for \(K_a\) giving approximately \( 1.07 \times 10^{-5}\).

Key concepts

Molar Mass Calculation

Calculating the molar mass of a substance allows us to understand its composition by determining how much one mole of that substance weighs. In this exercise, we're finding the molar mass of an unknown monoprotic acid. Start by calculating the moles of NaOH that reacted with the acid. You do this by using the formula for moles: \( ext{moles} = ext{molarity} imes ext{volume} \) in liters. Here, the volume of NaOH used is 18.4 mL, which we convert into liters by dividing by 1000.

• Moles of NaOH used: \( 0.0633 \; \text{M} \times \frac{18.4 \; \text{mL}}{1000} = 0.00116 \; \text{moles} \)

In a titration, the moles of the monoprotic acid react on a 1:1 basis with the moles of base. Thus, the moles of acid equals the moles of NaOH at the equivalence point.
To find the molar mass, divide the mass of the acid by the moles calculated:

• Molar mass of the acid: \( \frac{0.1276 \; \text{g}}{0.00116 \; \text{moles}} = 110 \; \text{g/mol} \)

This means that one mole of the unknown acid weighs 110 grams.

Acid Dissociation Constant

The acid dissociation constant, \( K_a \), is a crucial value that tells us how completely an acid dissociates in solution. It provides insight into the strength of an acid. In this problem, the pH after adding 10.0 mL of base is used to find \( K_a \) for the unknown acid.
First, acknowledge that a non-equivalence point involves only partial dissociation. Here, use the given pH to find \( [OH^-] \).

• The pH at this point is 5.87. Convert this to pOH with the relationship: \( \text{pOH} = 14 - \text{pH} = 8.13 \)
• Determine the concentration of hydroxide ions: \( [OH^-] = 10^{-\text{pOH}} = 10^{-8.13} = 7.4 \times 10^{-9} \; \text{M} \)

The concentrations of dissociated ions help solve for \( K_a \). Using the formula for weak acids:

• \( K_a = \frac{[H^+][A^-]}{[HA]} \)
• Assume \( [H^+] = [A^-] = x \), and substitute known values and assumptions to solve for \( K_a \).
• This yields approximately \( K_a = 1.07 \times 10^{-5} \).

A higher \( K_a \) value indicates a stronger acid, while a lower value signifies a weaker acid.

pH and pOH Relationship

The relationship between pH and pOH is essential in understanding chemical equilibria in acidic and basic solutions. They are both measures of proton (H+) and hydroxide (OH-) ion concentrations, respectively, in a solution.
In water at 25°C, the product of H+ and OH- concentrations is constant: \( [H^+][OH^-] = 1 \times 10^{-14} \). From this, we derive that:

• \( \text{pH} + \text{pOH} = 14 \)

This handy relationship allows us to easily convert between pH and pOH, provided one is known. It's essentially a reflection of the neutral water equilibrium scenario, where both ions are at equilibrium.
Understanding how these scales interconnect helps in many applications, such as titrations, where monitoring the pH change can indicate the completion of a reaction. Acids lower the pH, while bases lower the pOH, and vice versa. Thus, in any aqueous solution, knowing either pH or pOH gives you direct information about the other by subtraction from 14.
Mastering the use of this relationship is vital for solving problems that involve acid-base equilibria, such as titrations and buffer calculations. It ensures accurate calculations of concentrations, enabling prediction and control of reactions, especially during the titration process.