Problem 171 - A Complete Resource Book in Chemistry for JEE Main

Question

The \(\mathrm{pH}\) of a solution obtained by mixing equal volume of solutions having \(\mathrm{pH}=3\) and \(\mathrm{pH}=4 .[\log 5.5=\) \(0.7404]\) (a) \(3.26\) (b) \(3.5\) (c) \(4.0\) (d) \(3.42\)

Step-by-Step Solution

Answer

The pH of the mixed solution is 3.26.

Step 1: Understand the Problem

We need to calculate the pH of a solution created by mixing equal volumes of two solutions: one with a pH of 3 and the other with a pH of 4.

Step 2: Convert pH to \\[ ext{[H}^+ \\] ext{]}

Calculate the concentration of hydrogen ions for each solution using the formula \([ ext{H}^+]=10^{- ext{pH}}\). For pH 3, \[ ext{[H}^+] = 10^{-3}\]. For pH 4, \[ ext{[H}^+] = 10^{-4}\].

Step 3: Calculate the Average Concentration

When mixing equal volumes, we average the hydrogen ion concentrations. So, add them and divide by 2: \[\text{Average [H}^+] = \frac{10^{-3} + 10^{-4}}{2}\].

Step 4: Simplify the Average Concentration

Substitute the values: \[\text{Average [H}^+] = \frac{1 \, x\, 10^{-3} + 0.1 \, x\, 10^{-3}}{2} = \frac{1.1 \, x\, 10^{-3}}{2}\]. Then simplify this: \[0.55 \, x\, 10^{-3}\].

Step 5: Convert Back to pH

The pH is calculated from \[\text{pH} = -\log_{10}(\text{Average [H}^+] )\]. Plug in the average concentration: \[\text{pH} = -\log_{10}(0.55 \, x\, 10^{-3})\].

Step 6: Calculate the Final pH

By solving, \[\text{pH} = 3 - \log_{10}(0.55)\]. Given \[\log_{10} 5.5 = 0.7404\], we rearrange recognizing \[\log_{10}(0.55) = -1 + \log_{10}(5.5)\]. Thus, the pH = 3 - (0.7404 - 1) = 3.2596. Round it to two decimal places: 3.26.

Key Concepts

Hydrogen Ion Concentration

Understanding the concept of hydrogen ion concentration is foundational in acid-base chemistry. When we talk about the acidity of a solution, we refer to the number of free hydrogen ions, represented as \[\text{[H}^+]\]. This concentration determines how acidic a solution is. The lower the pH value, the higher the \[\text{[H}^+]\] and, consequently, the stronger the acidity.

• A solution with a pH of 3 means \[\text{[H}^+] = 10^{-3}\].
• For a pH of 4, \[\text{[H}^+] = 10^{-4}\].

These expressions show that slight changes in pH represent significant changes in hydrogen ion concentration, which explains why acids and bases can behave so differently with just minor pH adjustments.

Logarithmic Transformation

The logarithmic transformation is an essential mathematical tool used in chemistry to handle large ranges of data, especially in the context of pH calculations. The relationship between the pH and hydrogen ion concentration is logarithmic: \[\text{pH} = -\log_{10}(\text{[H}^+])\]. This transformation turns multiplication into addition or subtraction, simplifying the analysis of ratios.Say, for example, we have calculated an average hydrogen ion concentration and need to find the pH. We use the equation \[\text{pH} = -\log_{10}(0.55 \times 10^{-3})\]. Breaking it down:

• First manage the coefficient and power separately: \[\log_{10}(0.55 \times 10^{-3}) = \log_{10}(0.55) + \log_{10}(10^{-3})\].
• The latter simplifies to -3, while \(\log_{10}(0.55)\) can be derived from known values like \(\log_{10}(5.5) = 0.7404\) since \(\log_{10}(0.55) = \log_{10}(5.5) - 1\).
• The resulting calculation leads to the pH value, demonstrating how logarithms ease the division and multiplication in pH contexts.

Acid-Base Chemistry

Acid-base chemistry revolves around the concept of acids and bases, which are defined primarily by their ability to donate or accept hydrogen ions. In simple terms:

• Acids increase hydrogen ion concentration and thus lower the pH.
• Bases reduce hydrogen ion concentration, raising the pH.

When we mix solutions, as in the exercise, we're essentially combining different acid or base strengths. This requires careful balancing. Each solution contributes its own \[\text{[H}^+]\].By averaging these contributions, we find the overall effect on acidity or alkalinity in the mixed solution. Thus, achieving a new, combined pH. The beauty of acid-base chemistry lies in the simple yet profound interplay of hydrogen ions, making it a crucial part of understanding chemical solutions and reactions.