Problem 19 - General Chemistry: Principles and Modern Applications

Question

\(50.00 \mathrm{mL}\) of \(0.0155 \mathrm{M} \mathrm{HI}(\mathrm{aq})\) is mixed with \(75.00 \mathrm{mL}\) of 0.0106 M KOH(aq). What is the pH of the final solution?

Step-by-Step Solution

Answer

The pH of the final solution is 10.2

Step 1: Calculate the number of moles for HI and KOH

Using the formula Molarity (M)= moles of solute/liters of solution, calculate moles of HI and KOH. Moles of HI = Molarity * Volume = 0.0155 M * 0.050 L = \(7.75 \times 10^{-4} \, mol\). Moles of KOH = 0.0106 M * 0.075 L = \(7.95 \times 10^{-4} \, mol\)

Step 2: Determine the limiting reactant

In this case, HI is the limiting reactant and KOH is the excess reactant. This is because the moles of HI is less than the moles of KOH.

Step 3: Determine the number of remaining moles of KOH

Calculate the moles of KOH left after reaction with HI. This is obtained by subtracting the moles of HI from the moles of KOH: \(7.95 \times 10^{-4} mol \, - \, 7.75 \times 10^{-4} mol = 2 \times 10^{-5} mol\, KOH\)

Step 4: Calculate the OH- ion concentration in the solution

Since KOH is a strong base, it dissociates completely into K+ and OH- in solution. Therefore, the concentration of OH- ions is equal to the concentration of KOH. The volume of the solution is the sum of the volumes of HI and KOH, which is 0.050 L + 0.075 L = 0.125 L. The concentration \(OH^- = \frac { moles of OH^- }{Volume of Solution } = \frac {2 \times 10^{-5} mol} { 0.125 L} = 1.6 \times 10^{-4} M).

Step 5: Calculate the pOH and then the pH of the solution

The pOH is calculated as \( -\log[OH^-] \, = \, -\log(1.6 \times 10^{-4}) \, = \, 3.8 \). The pH is then calculated as \(14 - pOH \, = \, 14 - 3.8 = 10.2\)

Key Concepts

Molarity

Understanding molarity is crucial when dealing with concentrations in chemistry. Molarity, often symbolized as "M," refers to the concentration of a solution. Specifically, it defines how many moles of a solute are present in one liter of solution.
This concept is fundamental when calculating the amounts of substances involved in reactions. To calculate molarity, you use the formula:

• Molarity (M) = moles of solute / liters of solution

For example, if you have a solution with 0.0155 moles of hydrogen iodide (HI) in 1 liter, the molarity is 0.0155 M. If the same amount is in a 0.050-liter solution, as in the exercise, then it becomes 0.155 M due to the reduced volume. Understanding this concept allows students to predict how solutions will behave during chemical reactions.

Limiting Reactant

The limiting reactant in a chemical reaction is the substance that is completely consumed first, thus determining the extent of the reaction and limiting the amount of product formed.
In the exercise, when mixing HI and KOH, the reaction cannot continue once the HI is used up.

• To identify the limiting reactant: calculate the moles of each reactant.
• Compare these to determine which one runs out first.

For instance, in our case:

• Moles of HI = 0.0155 x 0.050 L = 0.000775 mol
• Moles of KOH = 0.0106 x 0.075 L = 0.000795 mol

Since the moles of HI are fewer than those of KOH, HI is the limiting reactant. Knowing this helps us determine how much KOH remains after the reaction ceases.

Acid-Base Reaction

An acid-base reaction is a chemical reaction that occurs between an acid and a base. It typically results in the formation of water and a new compound, often referred to as a salt.
Understanding these types of reactions is essential, particularly in solution chemistry.
In the practice exercise, HI acts as an acid, and KOH serves as a base. Here's what happens:

• HI donates a proton (H+) to OH- from KOH, generating water (H2O).
• The remaining ions (K+ and I-) form a neutral salt.

This process occurs in aqueous solutions and often results in an ionic equilibrium. Knowing how to calculate the products of these reactions can help determine aspects like the final pH of the solution, as acids and bases will impact the solution's acidity or basicity.

pOH and pH Relationship

pH and pOH are concepts that describe the acidity or basicity of a solution. They are inversely related on the pH scale, which ranges from 0 (strongly acidic) to 14 (strongly basic).
Knowing the relationship between pH and pOH is vital for calculating the concentration of hydrogen or hydroxide ions in a solution. Here’s how they relate:

• The pH is calculated using the formula: \( \text{pH} = -\log[\text{H}^+] \).
• The pOH is calculated as: \( \text{pOH} = -\log[\text{OH}^-] \).

Moreover, the sum of pH and pOH always equals 14 at 25°C:

• \( \text{pH} + \text{pOH} = 14 \)

In the given exercise, calculating the pOH from the concentration of OH- ions leads to finding the pH. Given the pOH of 3.8, the pH becomes 10.2, indicating a basic solution. Understanding this relationship lets you assess solution behavior and predict acidity or basicity after reactions.