Problem 279 - A Complete Resource Book in Chemistry for JEE Main

Question

How many litres of water must be added to litre of an aqueous solution of \(\mathrm{HCl}\) with a \(\mathrm{pH}\) of 1 to create an aqueous solution with \(\mathrm{pH}\) of \(2 ?\) (a) \(2.0 \mathrm{~L}\) (b) \(9.0 \mathrm{~L}\) (c) \(0.1 \mathrm{~L}\) (d) \(0.9 \mathrm{~L}\)

Step-by-Step Solution

Answer

You must add 9.0 L of water.

Step 1: Understand the pH Scale

The pH scale measures the acidity of a solution. A pH of 1 indicates a higher concentration of \( ext{H}^+\) ions than a pH of 2. Every point increase on the pH scale represents a tenfold decrease in \( ext{H}^+\) ion concentration.

Step 2: Calculate Initial \\(\text{H}^+\\) Concentration

The concentration of \(\text{H}^+\) ions in a solution with pH 1 is \[ \text{H}^+ = 10^{-\text{pH}} = 10^{-1} = 0.1 \, \text{M} \]

Step 3: Calculate Target \\(\text{H}^+\\) Concentration

The desired concentration of \(\text{H}^+\) ions in the solution with pH 2 is \[ \text{H}^+ = 10^{-\text{pH}} = 10^{-2} = 0.01 \, \text{M} \]

Step 4: Use Dilution Formula

Apply the dilution formula, which is \([C_1 \times V_1 = C_2 \times V_2]\), where \(C_1=0.1 \,\text{M}\), \(V_1=1 \,\text{L}\), and \(C_2=0.01 \,\text{M}\). We need to find \(V_2\): \[0.1 \times 1 = 0.01 \times V_2\] solving gives \[V_2 = 10 \,\text{L}\].

Step 5: Calculate Volume of Water to Add

To find out how much water to add, subtract the initial volume from the final volume: \[ V_{\text{water}} = V_2 - V_1 = 10 - 1 = 9 \, \text{L} \]

Key Concepts

Dilution Formula

When diluting a solution, the goal is to decrease the concentration of the solute, in this case, hydrogen ions from hydrochloric acid. The dilution formula is an essential tool in this context. It is represented as: \[ C_1 \times V_1 = C_2 \times V_2 \]where:

• \(C_1\) is the initial concentration of the solution.
• \(V_1\) is the initial volume of the solution.
• \(C_2\) is the desired concentration after dilution.
• \(V_2\) is the final volume after dilution.

This formula helps calculate the final volume needed to achieve a desired concentration. It is useful in many chemistry applications to adjust solution concentrations accurately. In the given exercise, you apply this formula to find the volume needed to dilute the acid solution from a pH of 1 to a pH of 2. Knowing this formula ensures precision when making solutions.

Hydrogen Ion Concentration

The concentration of hydrogen ions (\( \text{H}^+ \)) in a solution is a direct measure of its acidity. It's important to understand that the pH is a logarithmic scale, which means each step on this scale represents a tenfold change in \( \text{H}^+ \) concentration. For example, a solution with a pH of 1 has a \( \text{H}^+ \) concentration of 0.1 M, calculated as:\[ \text{H}^+ = 10^{-\text{pH}} = 10^{-1} = 0.1 \, \text{M} \]When the pH is increased to 2, the new concentration is:\[ \text{H}^+ = 10^{-2} = 0.01 \, \text{M} \]Understanding how to calculate these concentrations is pivotal for setting up reactions correctly. This knowledge allows you to manipulate and control the acidity of solutions, which is crucial in many scientific and industrial processes.

pH Scale

The pH scale is a vital tool in chemistry for gauging the acidity or basicity of solutions. It ranges from 0 to 14, where:

• A pH less than 7 indicates an acidic solution.
• A pH of 7 is neutral.
• A pH greater than 7 signifies a basic (alkaline) solution.

Specifically, a decrease in pH represents an increase in acidity. Each whole number change on the scale results in a tenfold change in hydrogen ion concentration. For instance, as seen in the exercise, a shift from a pH of 1 to a pH of 2 corresponds to a tenfold decrease in hydrogen ion concentration. This principle of the pH scale is instrumental for tasks ranging from adjusting the acidity of a swimming pool to formulating pharmaceuticals and understanding biochemical reactions in living organisms. Effective use of the pH scale helps ensure controlled and predictable chemical environments.