Problem 66 - Chemistry An Atoms First Approach

Question

In the titration of \(50.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) methylamine, \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) \(\left(K_{\mathrm{b}}=4.4 \times 10^{-4}\right),\) with \(0.50 M\) HCl, calculate the pH under the following conditions. a. after \(50.0 \mathrm{mL}\) of \(0.50 \mathrm{M}\) HCl has been added b. at the stoichiometric point

Step-by-Step Solution

Answer

a. After 50.0 mL of 0.50 M HCl has been added, the pH is approximately 10.64. b. At the stoichiometric point, the pH is approximately 10.12.

Step 1: a. After 50.0 mL of 0.50 M HCl has been added.

First, let's calculate the moles of methylamine and HCl in the solution: Moles of methylamine = volume × concentration = 50.0 mL × 1.0 M = 50.0 mmol Moles of HCl = volume × concentration = 50.0 mL × 0.50 M = 25.0 mmol Now, we need to determine the amount of CH3NH2 that has reacted with HCl and the amount of unreacted CH3NH2 and CH3NH3+ formed as a result of partial neutralization: Moles CH3NH2 reacted = 25.0 mmol Moles unreacted CH3NH2 = 50.0 mmol - 25.0 mmol = 25.0 mmol Moles of CH3NH3+ formed = moles of HCl added = 25.0 mmol Next, we determine the concentration of CH3NH2 and CH3NH3+ in the reaction mixture after adding 50.0 mL of HCl: Total volume = volume of methylamine solution + volume of HCl solution = 50.0 mL + 50.0 mL = 100 mL \[ CH\textsubscript{3}NH\textsubscript{2\, } = \frac{25\ mmol}{100\ mL} =0.25\ M \] \[ CH\textsubscript{3}NH\textsubscript{3\, }^\textsubscript{+} = \frac{25\ mmol}{100\ mL} =0.25\ M \] Now, using the Kb expression to find the hydroxide ion concentration OH-: Kb = 4.4 x 10^-4 = \[\frac{[OH^{-}][CH_{3}NH_{3}^{+}]}{[CH_{3}NH_{2}]}\] \[ [OH^{-}] = \frac{Kb [CH_{3}NH_{2}]}{[CH_{3}NH_{3}^{+}]} = \frac{4.4 \times 10^{-4} (0.25)}{0.25} = 4.4 \times 10^{-4}\] Now, we can calculate the pOH: \[pOH = -log_{10} [OH^{-}] = -log_{10}(4.4 \times 10^{-4}) \] \[pOH \approx 3.36\] Now, the pH can be calculated by subtracting the pOH from 14: \[ pH = 14 - pOH = 14 - 3.36 \] \[ pH \approx 10.64 \]

Step 2: b. At the stoichiometric point

At the stoichiometric point, equal moles of acid and base have reacted. Thus, Moles of HCl = Moles of CH3NH2 = 50.0 mmol To calculate the volume of HCl solution needed to react with all of the methylamine: Volume of HCl solution = \(\frac{moles\ of\ HCl}{concentration\ of\ HCl}\) = \(\frac{50.0\ mmol}{ 0.50\ M}\) = 100 mL At the stoichiometric point, the amount of CH3NH3+ formed is equal to the initial amount of CH3NH2: CH3NH3+ formed = moles of CH3NH2 = 50.0 mmol Now, we calculate the total volume at the stoichiometric point: Total volume = Volume of methylamine solution + Volume of HCl solution = 50.0 mL + 100 mL = 150 mL Thus, the concentration of CH3NH3+ is: \[CH\textsubscript{3}NH\textsubscript{3\, }^\textsubscript{+} = \frac{50.0\ mmol}{150\ mL} =0.333\ M \] We know that, Ka = Kw/Kb From that, we can calculate Ka for CH3NH3+: Ka = \(\frac{1.0 \times 10^{-14}}{4.4 \times 10^{-4}}\) ≈ 2.27 x 10^-11 Then, we can use the Ka expression to find the H+ concentration: \[ Ka = \frac{[H^{+}]}{[CH_{3}NH_{3}^{+}]} \] \[ [H^{+}] = Ka [CH_{3}NH_{3}^{+}] = 2.27 \times 10^{-11} \times 0.333 \] \[ [H^{+}] \approx 7.56 \times 10^{-11} \] Now, calculate the pH: \[ pH = -log_{10} [H^{+}] = -log_{10}(7.56 \times 10^{-11}) \] \[ pH \approx 10.12 \] So, the pH at the stoichiometric point is approximately 10.12.

Key Concepts

Methylamine

Methylamine, or \( \text{CH}_3\text{NH}_2 \), is a primary amine that acts as a weak base in aqueous solutions. It has a single methyl group (\( \text{CH}_3 \)) attached to an amine group (\( \text{NH}_2 \)), which alters its properties compared to a simple ammonia molecule. Due to this methyl group, methylamine is more basic than ammonia, making it more chemically reactive when mixed with acids like hydrochloric acid.

The basicity of methylamine is determined by its ability to accept a proton to form its conjugate acid, methylammonium, or \( \text{CH}_3\text{NH}_3^+ \). In titration, calculating the change in concentration of this species helps understand the progression of the reaction.

Key Properties of Methylamine:

• Acts as a proton acceptor in reactions.
• The presence of a methyl group increases base strength due to electron donation.
• It forms a conjugate acid, methylammonium, during neutralization.

Stoichiometric Point

The stoichiometric point in a titration marks the moment when the amount of titrant added is chemically equivalent to the amount of substance present in the solution. For the titration of methylamine with hydrochloric acid, this point is reached when all the methylamine has been converted into its conjugate acid: methylammonium.

Achieving this point means the reaction between the methylamine and hydrochloric acid is complete, and this is crucial for proper pH calculation. At the stoichiometric point, no "extra" acid or base remains to further alter the pH directly.

**Characteristics of the Stoichiometric Point:**

• All original base molecules react completely to form the conjugate acid.
• Volume and concentration of the acid added should equal the moles of the base present initially.
• The pH reflects the characteristics of the conjugate acid, often leading to a theoretical pH calculation based on weak acid behavior.

Hydrochloric Acid

Hydrochloric acid, also known as HCl, is a strong monobasic acid often used in titration experiments. It fully dissociates in an aqueous solution, releasing hydrogen ions (\( \text{H}^+ \)) and chloride ions (\( \text{Cl}^- \)). These free hydrogen ions engage with bases in neutralization reactions, such as the reaction with methylamine in titration.

When conducting a titration of a base like methylamine with HCl, it’s vital to measure accurately to precisely reach the stoichiometric point. At this point, the solution is a mix primarily of the resulting methylammonium and chloride ions.

• Properties of HCl in Titration:
  • Highly dissociative, contributing protons readily.
  • Used to standardize bases during titration due to its strong acidic nature.
  • Combined with bases to form salt and water in neutralization reactions.

pH Calculation

Calculating pH in titration experiments involves understanding the relationship between hydrogen and hydroxide ions within the solution. For methylamine, after neutralization, knowing the concentrations of the remaining methylamine and methylammonium allows the use of the equilibrium expressions to determine the relevant ion concentrations.

In a titration involving weak bases and strong acids, the pH is calculated from the concentration of hydrogen ions (\( \text{H}^+ \)), derived from the strength of the conjugated forms reacting together.

To calculate pH during methylamine titration:

• The neutralization equation is applied to find the moles of remaining substances.
• The total reaction volume informs the final concentration of ions.
• Use the formula \(\text{pH} = 14 - \text{pOH} \) to find the pH from the hydroxide ion concentration.

In summary, by carefully following the mathematical derivations and considering equilibrium constants, accurate pH outcomes can be achieved for the solution at and around the stoichiometric point.