Problem 71 - General Chemistry: Principles and Modern Applications

Question

You are asked to bring the \(\mathrm{pH}\) of \(0.500 \mathrm{L}\) of \(0.500 \mathrm{M}\) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})\) to 7.00 How many drops \((1 \text { drop }=0.05 \mathrm{mL})\) of which of the following solutions would you use: \(10.0 \mathrm{M} \mathrm{HCl}\) or \(10.0 \mathrm{M} \mathrm{NH}_{3} ?\)

Step-by-Step Solution

Answer

Approximately 496 drops of \(10.0 M NH_3\) solution are needed to bring the pH of the \(0.500 L\) of \(0.500 M NH_4Cl\) solution to 7.00.

Step 1: Compute the initial moles of \(NH_4Cl\)

First, find the number of moles of \(NH_4Cl\) present in the 0.500 L solution by multiplying its volume by its molarity: \(0.500L \times 0.500mol/L = 0.250mol\) of \(NH_4Cl\).

Step 2: Calculate the \(Kb\) of \(NH_3\)

Next, using the \(Kw\) (\(1.00 \times 10^{-14}\)) and \(Ka\) (\(5.56 \times 10^{-10}\)) values for \(NH_4^+\), calculate the \(Kb\) for \(NH_3\) with the formula \[ Kb = Kw / Ka \] This gives \(Kb = (1.00 \times 10^{-14}) / (5.56 \times 10^{-10}) ≈ 1.80 \times 10^{-5}\].

Step 3: Create an ICE table

The chemical equation for the dissociation of \(NH_4Cl\) is \[ NH_4^+ + H2O ⇌ NH3 + H3O^+ \] The ICE table would look like this: \nI 0.250 - 0 - \nC -x - +x +x \nE 0.250-x - x x \nKb = [ \(NH_3\) ][ \(H_3O^+\) ] / [ \(NH_4^+\) ] \nplug in values and solve for x \n\(1.80 \times 10^{-5}\) = (x)(x) / (0.250 - x)\n\nAssuming x << 0.250, the equation simplifies to \[ x = √[(0.250) (1.80 \times 10^{-5})] = 2.12 \times 10^{-3} M \] The concentration of \(NH_3\) needed to bring pH to 7.00 is \(2.12 \times 10^{-3} M\).

Step 4: Determine which solution is needed

Since the \(NH_3\) concentration is lower than \(NH_4Cl\) concentration, a base needs to be added. Thus, drops of \(10.0 M NH_3\) solution should be added.

Step 5: Calculate the number of drops

Finally, calculate the number of moles of \(NH_3\) needed, which is equal to the moles of \(NH_4Cl\), using the formula: \[ molesNH_3 = molesNH_{4}Cl - [NH_3] = 0.250mol - 2.12 \times 10^{-3} mol ≈ 0.248mol \] To find the volume of \(10.0 M NH_3\) needed, divide the moles of \(NH_3\) by its molarity, which gives 0.0248 L or 24.8 mL. Now, to get the number of drops, divide 24.8 mL by the volume per drop (0.05 mL/drop), which equals to about 496 drops.

Key Concepts

pH calculation

Understanding pH calculation is essential when dealing with buffer solutions. The pH of a solution denotes its acidity or basicity, ranging from 0 (very acidic) to 14 (very basic). Calculating pH requires knowledge of the concentrations of acidic and basic components in the solution.
For instance, in the exercise, we aimed to achieve a pH of 7.00, which is neutral. A pH of 7 indicates an equal concentration of hydrogen ions (\(H^+\)) and hydroxide ions (\(OH^-\)) in the solution.
In a mathematical context, pH is calculated using the formula: \[\text{pH} = -\log_{10}([H^+])\] However, when adjusting the pH of a buffer solution, we typically focus on the interaction between the acid and its conjugate base. This requires identifying the concentrations that will yield the desired pH.

• For basic solutions, as in the original problem, account for the concentrations of the base (\(NH_3\)) and its conjugate acid (\(NH_4^+\)).
• Calculate the pH considering the equilibrium established between them.
• Determine the necessary adjustments by adding either an acid or a base to achieve the target pH.

acid-base equilibrium

Acid-base equilibrium involves the balance between concentrations of acids and bases in a chemical reaction. This concept is pivotal when working with buffer solutions, which resist changes in pH upon the addition of small amounts of acid or base.
In the given problem, \(NH_4Cl\) dissociates in water to form \(NH_4^+\) and \(Cl^-\). The \(NH_4^+\) further establishes an equilibrium with \(NH_3\) and \(H_3O^+\) ions in solution, a key aspect of acid-base reactions.
This equilibrium can be represented by the equation: \[ NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+ \] Using an ICE (Initial, Change, Equilibrium) table, we can explore how concentrations change at equilibrium. This approach helps us understand that when \(x\) changes, it impacts the equilibrium state.
The equilibrium constant expression, facilitated by the base dissociation constant (\(K_b\)), provides a quantitative measure of equilibrium for weak bases like \(NH_3\). This constant is essential for calculating the concentration of components in solution and adjusting the solution to reach desired pH levels.

molarity

Molarity is a measure of the concentration of a solute in a solution. It is vital for calculations in any chemistry problem involving solutions. Molarity (\(M\)), expressed in moles per liter (\(mol/L\)), helps us understand the amount of a substance in a given volume.
For example, with \(NH_4Cl\), the molarity was 0.500 M, which indicates there are 0.500 moles of \(NH_4Cl\) per liter of solution. This quantity is crucial when determining the initial moles of the solute.

• To calculate moles from molarity, multiply the molarity by the volume of the solution (\(0.500 \, L \, \times \, 0.500 \, M = 0.250 \, moles\) of \(NH_4Cl\)).
• Understanding molarity is also key when deciding how much \(NH_3\) to add; using the molarity of \(10.0 \, M\) \(NH_3\), we calculated the number of drops needed to maintain the buffer system close to the desired pH.
• The molarity indicates how diluted or concentrated a solution is, thus influencing how much of a reactant should be added to achieve a target pH.

Determining these factors is critical for precise adjustments in chemical reactions.