Problem 44 - Chemistry: An Atoms-Focused Approach

Question

A \(25.0 \mathrm{mL}\) sample of a \(0.100 M\) solution of aqueous trimethylamine is titrated with a \(0.125 M\) solution of \(\mathrm{HCl}\) at \(25^{\circ} \mathrm{C} .\) What is the pH of the solution after 10.0,20.0 and 30.0 mL of acid have been added?

Step-by-Step Solution

Answer

Answer: The pH values after titrating with 10.0, 20.0, and 30.0 mL of HCl are 12.52, 12.70, and 1.60, respectively.

Step 1: Calculate the initial moles of trimethylamine

To calculate the moles of trimethylamine, we use the formula: Moles of \(\mathrm{N(CH_3)_3} = \) volume × concentration Moles of \(\mathrm{N(CH_3)_3} = 25.0\,\mathrm{mL} \times 0.100\,\mathrm{M}\) Moles of \(\mathrm{N(CH_3)_3} = 0.00250\,\mathrm{mol}\)

Step 2: Calculate moles of HCl added

For each titration point, we calculate the moles of \(\mathrm{HCl}\) added using the formula: Moles of \(\mathrm{HCl} = \) volume × concentration Moles of \(\mathrm{HCl}\) added after \(10.0\,\mathrm{mL}\) of titration: \(= 10.0\,\mathrm{mL} \times 0.125\,\mathrm{M} = 0.00125\,\mathrm{mol}\) Moles of \(\mathrm{HCl}\) added after \(20.0\,\mathrm{mL}\) of titration: \(= 20.0\,\mathrm{mL} \times 0.125\,\mathrm{M} = 0.00250\,\mathrm{mol}\) Moles of \(\mathrm{HCl}\) added after \(30.0\,\mathrm{mL}\) of titration: \(= 30.0\,\mathrm{mL} \times 0.125\,\mathrm{M} = 0.00375\,\mathrm{mol}\)

Step 3: Determine moles of trimethylamine and N(CH_3)_3H^+

At each titration point, we have the following: After \(10.0\,\mathrm{mL}\) of titration: Moles of \(\mathrm{N(CH_3)_3} = 0.00250\,\mathrm{mol} - 0.00125\,\mathrm{mol} = 0.00125\,\mathrm{mol}\) Moles of \(\mathrm{N(CH_3)_3H^+} = 0.00125\,\mathrm{mol}\) After \(20.0\,\mathrm{mL}\) of titration: Moles of \(\mathrm{N(CH_3)_3} = 0.00250\,\mathrm{mol} - 0.00250\,\mathrm{mol} = 0\,\mathrm{mol}\) Moles of \(\mathrm{N(CH_3)_3H^+} = 0.00250\,\mathrm{mol}\) After \(30.0\,\mathrm{mL}\) of titration: At this point, there is an excess of \(\mathrm{HCl}\), so the moles of \(\mathrm{N(CH_3)_3}\) remain at \(0\,\mathrm{mol}\). Moles of excess \(\mathrm{HCl} = 0.00375\,\mathrm{mol} - 0.00250\,\mathrm{mol} = 0.00125\,\mathrm{mol}\)

Step 4: Calculate the concentration of N(CH_3)_3H^+

At each titration point, we have the following: After \(10.0\,\mathrm{mL}\) of titration: Concentration of \(\mathrm{N(CH_3)_3H^+} = \frac{0.00125\,\mathrm{mol}}{25.0\,\mathrm{mL} + 10.0\,\mathrm{mL}} = 0.0333\,\mathrm{M}\) After \(20.0\,\mathrm{mL}\) of titration: Concentration of \(\mathrm{N(CH_3)_3H^+} = \frac{0.00250\,\mathrm{mol}}{25.0\,\mathrm{mL} + 20.0\,\mathrm{mL}} = 0.0500\,\mathrm{M}\) After \(30.0\,\mathrm{mL}\) of titration (excess \(\mathrm{HCl}\)): Concentration of \(\mathrm{HCl} = \frac{0.00125\,\mathrm{mol}}{25.0\,\mathrm{mL} + 30.0\,\mathrm{mL}} = 0.0250\,\mathrm{M}\)

Step 5: Calculate the pH

At each titration point, we have the following: After \(10.0\,\mathrm{mL}\) of titration: \(p\mathrm{OH} = -\log{\left(0.0333\right)} = 1.48\) \(p\mathrm{H} = 14 - p\mathrm{OH} = 14 - 1.48 = 12.52\) After \(20.0\,\mathrm{mL}\) of titration: \(p\mathrm{OH} = -\log{\left(0.0500\right)} = 1.30\) \(p\mathrm{H} = 14 - p\mathrm{OH} = 14 - 1.30 = 12.70\) After \(30.0\,\mathrm{mL}\) of titration (excess \(\mathrm{HCl}\)): \(p\mathrm{H} = -\log{\left(0.0250\right)} = 1.60\) Therefore, the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added are 12.52, 12.70, and 1.60, respectively.

Key Concepts

pH calculation

Calculating the pH of a solution is essential for understanding the acidic or basic nature of that solution.

The pH is determined through the relationship with hydrogen ion concentration: \[ pH = -\log{[H^+]} \] The pH scale ranges from 0 to 14, where a pH less than 7 indicates an acidic solution, exactly 7 is neutral, and greater than 7 is basic.

When dealing with a base such as trimethylamine, we first find the pOH using the hydroxide ion concentration:\[ pOH = -\log{[OH^-]} \]Finally, by using the relationship between pH and pOH, which is the constant across any aqueous solution at 25°C:\[ pH + pOH = 14 \]we can determine the pH. This calculation becomes more involved during titration as acids are added, changing the concentrations of ions involved.

acid-base reaction

An acid-base reaction involves the transfer of protons (hydrogen ions) from the acid to the base. In a titration scenario, such as this experiment,

an acid (HCl) is gradually added to a base solution (trimethylamine). Initially, trimethylamine (\( \text{N(CH}_3\text{)}_3 \) ) acts as a base and accepts protons from the \( \text{HCl} \), forming \( \text{NH(CH}_3\text{)}_3^+ \).

As the titration progresses, more HCl reduces the concentration of trimethylamine while increasing the concentration of the conjugate acid, \( \text{N(CH}_3\text{)}_3^+ \).The equivalence point in a titration is reached when all the base has reacted with the added acid, resulting in the formation of water and a salt, indicating that the solution will either be neutral or, as in some cases like this with a strong acid and weak base, slightly acidic depending on the remaining species in solution.

chemical equilibrium

Chemical equilibrium occurs in a reversible reaction when the rate of the forward reaction equals the rate of the reverse reaction, meaning the concentrations of reactants and products remain constant throughout.

In the titration of trimethylamine and hydrochloric acid, equilibrium is significant when evaluating the protonation of \( \text{N(CH}_3\text{)}_3 \) by \( \text{HCl} \), to form \( \text{NH(CH}_3\text{)}_3^+ \).This equilibrium shifts as varying quantities of \( \text{HCl} \) are added.

Before the equivalence point, there's usually a dynamic balance between \( \text{N(CH}_3\text{)}_3 \) and its protonated form. Once all the \( \text{N(CH}_3\text{)}_3 \) molecules are protonated, if more \( \text{HCl} \) is added, it disrupts the equilibrium fully to the side of products. This is why the pH can drop significantly once all the base is neutralized.

molar concentration

Molar concentration, often denoted as M, describes the concentration of a substance in a solution.

It is defined as the number of moles of solute per liter of solution:

\[ C = \frac{n}{V} \]

• C: Molar concentration in moles per liter (M)
• n: Number of moles of solute
• V: Volume of the solution in liters

In the titration context, understanding the molar concentration is crucial.

In the given problem, the concentration for both the \( \text{N(CH}_3\text{)}_3 \) and \( \text{HCl} \) determines how much each reactant is added or has reacted at various points in the titration.By multiplying the concentration by the volume of the solution, you get the number of moles, which is essential for stoichiometry during titration and pH calculation.