Problem 44 - Precalculus with Calculus Previews

Question

Determine the \(\mathrm{pH}\) of a solution with the given hydrogen-ion concentration \(\left[\mathrm{H}^{+}\right]\). $$ 5.1 \times 10^{-5} $$

Step-by-Step Solution

Answer

The pH of the solution is approximately 4.292.

Step 1: Understand the pH formula

The pH of a solution is calculated using the formula \[ \text{pH} = -\log_{10} [\text{H}^+] \] where \([\text{H}^+]\) is the concentration of hydrogen ions in moles per liter.

Step 2: Substitute the given concentration

Substitute the given hydrogen-ion concentration into the pH formula. The concentration provided is \( [\text{H}^+] = 5.1 \times 10^{-5} \). So, the formula becomes \[ \text{pH} = -\log_{10} (5.1 \times 10^{-5}) \].

Step 3: Calculate the pH using logarithm rules

To solve \( \text{pH} = -\log_{10} (5.1 \times 10^{-5}) \), remember the logarithm rule: \( \log_{10}(a \times b) = \log_{10} a + \log_{10} b \). Thus, \[ \text{pH} = - (\log_{10} 5.1 + \log_{10} 10^{-5}) \].

Step 4: Simplify the expression using logarithm values

First, calculate \( \log_{10} 10^{-5} = -5 \). Then estimate \( \log_{10} 5.1 \). Using a calculator, \( \log_{10} 5.1 \approx 0.708 \). Substituting these into the expression gives \[ \text{pH} = - (0.708 - 5) = 4.292 \].

Step 5: Finalize the solution

Therefore, the pH of the solution, after evaluating the expression, is approximately \(4.292\).

Key Concepts

Logarithms

Logarithms are incredibly important in various mathematical scenarios, especially in chemistry where we often deal with tiny numbers like hydrogen-ion concentrations. In essence, a logarithm in base 10, denoted as \( \log_{10} \), helps us to find the exponent to which 10 must be raised to obtain a given number. For instance, the logarithm of 1000 is 3 because \(10^3 = 1000\).

This concept comes in handy when dealing with pH calculations, as it allows us to operate on hydrogen ion concentrations directly. The formula for pH is \( \text{pH} = -\log_{10} [\text{H}^+] \), which translates these concentrations into manageable numbers. Instead of working with very small numbers like \( 5.1 \times 10^{-5} \), we can use logarithms to make these figures more interpretable. This approach simplifies calculations and is widely used in various scientific fields. Remember the key rule that \( \log_{10}(a \times b) = \log_{10} a + \log_{10} b \). Knowledge of logarithms makes the process not just easier, but also more intuitive.

Hydrogen-Ion Concentration

The concentration of hydrogen ions \((\text{H}^+)\) in a solution is a crucial factor in chemistry because it determines the acidity or basicity of that solution. In the exercise provided, we have a concentration of \( [\text{H}^+] = 5.1 \times 10^{-5} \) moles per liter.

This small number reflects the great sensitivity and range that the pH scale covers. High concentrations of \( \text{H}^+ \) ions mean a low pH, indicating a more acidic solution, while low concentrations correspond to a higher pH and thus a basic solution.

• pH scale ranges from 0 to 14.
• pH less than 7 indicates acidity.
• pH greater than 7 indicates basicity.
• pH of 7 is neutral, like pure water.

By converting \( [\text{H}^+] \) to a pH value, we can easily compare concentrations and make quantitative assessments of acidity. This is why the logarithmic pH calculation is a practical tool in chemistry.

Chemistry Education

Chemistry education aims to provide students with a solid understanding of fundamental concepts that explain how substances interact. pH calculations are foundational because they help explain how substances behave in different environments.

Understanding the concept of logarithms and hydrogen-ion concentrations is a stepping stone to grasping more complex chemical interactions. These topics teach students to:

• Recognize the importance of precision in measurements.
• Use scientific notation to describe very large or very small numbers.
• Apply algebraic rules, like logarithm rules, in practical scenarios.
• Relate mathematical formulas to real-world chemistry problems.

Developing competence in these areas is crucial for more advanced chemistry topics like buffers, titrations, and reaction kinetics. By breaking these concepts down, educators can make the subject more approachable and stimulate curiosity in learners.