Logarithms and Their Applications in GOB Chemistry

Logarithms in Chemistry

Introduction to Logarithms

Logarithms are mathematical functions that are widely used in chemistry, especially in calculations involving pH, reaction rates, and equilibrium. The logarithmic base 10 (common logarithm) and the natural logarithm (base e) are particularly important in General, Organic, and Biological (GOB) Chemistry.

• Logarithm (base 10): The exponent to which 10 must be raised to obtain a given number.

• Natural logarithm (ln): The exponent to which the base e (approximately 2.718) must be raised to obtain a given number.

Logarithmic Functions and Their Properties

Base 10 Logarithms

The base 10 logarithm, written as log, is defined as follows:

• If $10^x = y$, then $\log y = x$.

• Examples:

  • $10^1 = 10$ so $\log 10 = 1$

  • $10^4 = 10,000$ so $\log 10,000 = 4$

  • $10^0 = 1$ so $\log 1 = 0$

  • $10^{-2} = 0.01$ so $\log 0.01 = -2$

Inverse Logarithmic Functions

The inverse of a logarithmic function is called the antilogarithm:

• If $\log x = y$, then $x = 10^y$.

• Examples:

  • $\log 100 = 2$ so $\text{inv log } 2 = 10^2 = 100$

  • $\log 0.01 = -2$ so $\text{inv log } -2 = 10^{-2} = 0.01$

Natural Logarithms (ln)

The natural logarithm uses base e:

• If $e^x = y$, then $\ln y = x$.

• Examples:

  • $\ln 1000 = 6.908$ (since $e^{6.908} \approx 1000$)

  • $\ln 2 = 0.693$ (since $e^{0.693} \approx 2$)

• The inverse: $\text{inv ln } y = e^y$

Logarithmic Calculations: Examples

Example 1: Calculating Logarithms Without a Calculator

• $\log (1.0 \times 10^{-7}) = \log 1.0 + \log (10^{-7}) = 0 + (-7) = -7$

• $\log 1000 = \log (10^3) = 3$

Example 2: More Practice

• $\log (1.0 \times 10^5) = \log 1.0 + \log (10^5) = 0 + 5 = 5$

• $\log 0.0001 = \log (10^{-4}) = -4$

Example 3: Inverse Logarithms

• $\text{inv log } 2 = 10^2 = 100$

• $\text{inv log } 0.693 = 10^{0.693} \approx 4.93$

• $\text{inv ln } 0.693 = e^{0.693} = 2$

Logarithms in Buffer Calculations

The Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is used to calculate the pH of a buffer solution:

• $\text{pH} = \text{p}K_a + \log \left( \frac{\text{conjugate base}}{\text{weak acid}} \right)$

• Given pH = 4.17 and pKa = 3.83, solve for the ratio:

• $4.17 = 3.83 + \log \left( \frac{\text{conjugate base}}{\text{weak acid}} \right)$

• $\log \left( \frac{\text{conjugate base}}{\text{weak acid}} \right) = 4.17 - 3.83 = 0.34$

• $\frac{\text{conjugate base}}{\text{weak acid}} = 10^{0.34} \approx 2.19$

Natural Logarithms in Chemistry

Properties and Examples

• $\ln x = y$ means $e^y = x$

• Example: $\ln 1000 = 6.908$

• Inverse: $\text{inv ln } 6.908 = e^{6.908} = 1000$

Example: Solving for a Variable Using Natural Logarithms

• Given: $\ln[x] = -(2.13 \times 10^{-1})(12.3) + \ln[1.25]$

• Calculate $-(2.13 \times 10^{-1})(12.3) = -2.6199$

• $\ln[x] = -2.6199 + \ln[1.25]$

• $\ln[1.25] = 0.2231$

• $\ln[x] = -2.6199 + 0.2231 = -2.3968$

• $x = e^{-2.3968} \approx 0.091$

Logarithmic Properties and Rules

Key Logarithmic Relationships

Operation	Base 10 Logarithm	Natural Logarithm
Multiplication	$\log(a \cdot b) = \log a + \log b$	$\ln(a \cdot b) = \ln a + \ln b$
Division	$\log \left( \frac{a}{b} \right) = \log a - \log b$	$\ln \left( \frac{a}{b} \right) = \ln a - \ln b$
Power	$\log(a^x) = x \log a$	$\ln(a^x) = x \ln a$
Root	$\log(\sqrt[n]{a}) = \frac{1}{n} \log a$	$\ln(\sqrt[n]{a}) = \frac{1}{n} \ln a$

Example: Applying Logarithmic Properties

• If $\log 3 = 0.48$ and $\log 2 = 0.30$, what is $\log 12$?

• $12 = 3 \times 2 \times 2$ so $\log 12 = \log 3 + 2 \log 2 = 0.48 + 2(0.30) = 1.08$

Summary Table: Logarithmic and Exponential Relationships

Expression	Equivalent Form
$10^x = y$	$\log y = x$
$e^x = y$	$\ln y = x$
$\log x = y$	$x = 10^y$
$\ln x = y$	$x = e^y$

Additional info: Logarithms are essential in GOB Chemistry for understanding pH, buffer systems, and reaction kinetics. Mastery of logarithmic properties allows for efficient problem-solving in these areas.