Problem 88 - General Chemistry: Principles and Modern Applications

Question

An aqueous solution has \(109.2 \mathrm{g} \mathrm{KOH} / \mathrm{L}\) solution. The solution density is \(1.09 \mathrm{g} / \mathrm{mL} .\) Your task is to use \(100.0 \mathrm{mL}\) of this solution to prepare \(0.250 \mathrm{m}\) KOH. What mass of which component, \(\mathrm{KOH}\) or \(\mathrm{H}_{2} \mathrm{O}\), would you add to the \(100.0 \mathrm{mL}\) of solution?

Step-by-Step Solution

Answer

9.517 grams of water (H2O) need to be added to the 100 mL of solution to prepare 0.250 M KOH.

Step 1: Find mass of KOH in 100ml solution

1. First, convert the volume of the solution from ml to L by dividing by 1000, which gives \(0.1 \mathrm{L}\). Then, \nThe mass of KOH in the 100 ml solution can be calculated using the formula: \nMass = Volume (L) × Concentration (g/L) = \(0.1 \mathrm{L} × 109.2 \mathrm{g}/\mathrm{L} = 10.92 \mathrm{g} \)

Step 2: Calculate the mass of KOH needed in desired solution

To find the mass of KOH required for a 0.250 M KOH solution in 100ml (or 0.1 L), we can use the formula: \nMass = Molarity (M) × Volume (L) × Molar Mass (g/mol) \nAs the molar mass of KOH is approximately 56.11 g/mol, we find \nMass = \(0.250\, \mathrm{M}\, ×\, 0.1\, \mathrm{L}\, ×\, 56.11\, \mathrm{g/mol}\) = 1.403 \mathrm{g} \nThis is the mass of KOH we need in our 0.250 M KOH solution.

Step 3: Find the mass of the component to be added

Subtract the mass of KOH in desired solution found in step 2 from mass of KOH in the initial solution found in step 1. So, the mass difference is \(10.92 \mathrm{g} - 1.403 \mathrm{g} = 9.517 \mathrm{g}\). Since we get a positive number, this means that we have excess KOH in our solution. As a result, we would need to add water (H2O), not KOH to our solution to achieve the desired molarity.

Key Concepts

KOH Solution

When we talk about KOH solutions, we're referring to potassium hydroxide dissolved in water to form an aqueous solution. KOH, known as caustic potash, is commonly used in laboratories and industries due to its strong alkaline properties.

A KOH solution is often characterized by its concentration, which tells us how much KOH is present in a certain volume of solution. In our exercise, the solution has a concentration of 109.2 g/L, meaning there are 109.2 grams of KOH dissolved in every liter of water. This high concentration indicates that the solution is quite strong, capable of reacting significantly with other substances.

Understanding the nature and behavior of KOH solutions is crucial when performing various chemical reactions and processes. Knowing how to properly dilute and prepare solutions with specific molarities is an essential skill in chemistry.

Concentration Calculations

Concentration calculations are fundamental in chemistry and help us determine how much of a substance we have in a given solution. In practical terms, it allows us to know how strong or weak a solution is. This is particularly important for reactions where the quantity of reactants must be precise.

In the original exercise, to find how much KOH is in 100 mL of solution, we use the concentration (109.2 g/L) converted to match the smaller volume:

• First, convert 100 mL to liters by dividing by 1000, resulting in 0.1 L.
• Multiply this volume by the concentration: 0.1 L × 109.2 g/L = 10.92 g of KOH.

This calculation tells us that there are 10.92 grams of KOH in 100 mL of the solution.

By applying concentration calculations, we are better equipped to modify and tailor solutions to meet specific experimental requirements. Such calculations are vital both in laboratories and industrial applications.

Molarity

Molarity is a way of expressing concentration, specifically noting the number of moles of a solute in one liter of solution. It is denoted by the symbol 'M' and is one of the most widely used concentration units in chemistry.

To calculate the molarity of a solution, you need to know the amount of solute (in moles) and the volume of the solution (in liters). In our exercise, we aim to prepare a 0.250 M KOH solution. Utilizing the formula \( \text{Mass} = \text{Molarity (M)} \times \text{Volume (L)} \times \text{Molar Mass (g/mol)} \), we find out how much KOH is needed for a given solution volume. Here, the molar mass of KOH is 56.11 g/mol. By inserting the known values: \( 0.250 \, \text{M} \times 0.1 \, \text{L} \times 56.11 \frac{\text{g}}{\text{mol}} = 1.403 \, \text{g} \).

This implies that to prepare a 0.250 M solution, we require 1.403 g of KOH in 0.1 L of solution. Understanding molarity and mastering the calculations enable chemists to create solutions accurately and effectively, thereby playing a crucial role in scientific research and chemical manufacturing.