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Chapter 12: Problem 34

Find the pH of the substance with the given hydronium ion concentration. Sodium bicarbonate, \(4.0 \times 10^{-9}\)

Short Answer

Expert verified
The pH of the sodium bicarbonate solution is approximately 8.40.

Step by step solution

01

- Understand the pH formula

The pH of a substance can be found using the formula: \( \text{pH} = -\text{log}[\text{H}^+] \), where \( [\text{H}^+] \) is the hydronium ion concentration.
02

- Identify the given concentration

The given hydronium ion concentration for sodium bicarbonate is \( 4.0 \times 10^{-9} \).
03

- Substitute the concentration into the pH formula

Substitute \( 4.0 \times 10^{-9} \) into the formula: \( \text{pH} = -\text{log}(4.0 \times 10^{-9}) \).
04

- Compute the logarithm

Calculate the logarithm: \( \text{log}(4.0 \times 10^{-9}) = \text{log}(4.0) + \text{log}(10^{-9}) = 0.60206 - 9 = -8.39794 \).
05

- Calculate the pH

Finally, apply the negative sign to the logarithm result: \( \text{pH} = -(-8.39794) = 8.39794 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydronium Ion Concentration
The hydronium ion concentration \[ [\text{H}_3\text{O}^+] \] refers to the number of hydronium ions present in a solution. Hydronium ions form when an acid dissolves in water, releasing hydrogen ions \[ (\text{H}^+) \] which then combine with water molecules \[ (\text{H}_2\text{O}) \] to form \[ \text{H}_3\text{O}^+ \]. This concentration gives us an idea of how acidic or basic a solution is.
In this example, sodium bicarbonate has a hydronium ion concentration of \[ 4.0 \times 10^{-9} \], which indicates it is a weak base.
  • Lower concentrations of hydronium ions indicate a higher pH (basic solution).
  • Higher concentrations of hydronium ions indicate a lower pH (acidic solution).
Understanding hydronium ion concentration is key for calculating the pH of any solution.
pH Formula
The pH formula is essential in determining how acidic or basic a solution is. The pH is calculated using: \[ \text{pH} = -\text{log}[\text{H}^+] \], where \[ [\text{H}^+] \] is the concentration of hydronium ions in the solution.
For instance, we are given a hydronium ion concentration of \[ 4.0 \times 10^{-9} \] M for sodium bicarbonate. Plugging this into the formula:
  • \[ \text{pH} = -\text{log}(4.0 \times 10^{-9}) \]
  • Applying logarithm rules, we split it into two parts: \[ \text{log}(4.0) + \text{log}(10^{-9}) \]
Once we understand this, our next steps involve calculating the logarithms accurately.
Logarithm in Chemistry
Using a logarithm function in chemistry is common for dealing with a wide range of concentration values. A logarithm (log) helps transform large variations in values into more manageable numbers.
For pH calculation:
  • \[ \text{log}(4.0 \times 10^{-9}) = \text{log}(4.0) + \text{log}(10^{-9}) = 0.60206 - 9 \]
  • Here, \[ \text{log}(4.0) \] is approximately 0.60206 and \[ \text{log}(10^{-9}) = -9 \].
  • Combining these gives us: \[ \text{log}(4.0 \times 10^{-9}) = -8.39794 \]
By understanding how to apply logarithms, we can simplify complex multiplications and divisions often found in chemical calculations.
Acid-Base Chemistry
Acid-base chemistry explores the properties and reactions of acids and bases. An acid increases the hydronium ion concentration in a solution, while a base decreases it by increasing the hydroxide ion concentration.
The pH scale ranges from 0 to 14 and provides a measure of acidity or basicity:
  • pH < 7: Acidic solution
  • pH = 7: Neutral solution
  • pH > 7: Basic solution
In this example, sodium bicarbonate has a pH of 8.39794, making it a weak base. Understanding the pH and properties of acids and bases helps us predict reaction behavior and acidity levels in various solutions.

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Most popular questions from this chapter

Find logarithm. Give approximations to four decimal places. \(\log 43\)

Solve equation. \(\log _{a} 1=0\)

Find each logarithm. Give approximations to four decimal places. \(\log \left(4.76 \times 10^{9}\right)\)

The value of \(e\) can be expressed as $$e=1+\frac{1}{1}+\frac{1}{1 \cdot 2}+\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{1 \cdot 2 \cdot 3 \cdot 4}+\cdots$$ Approximate \(e\) using two terms of this expression, then three terms, four terms, five terms, and six terms. How close is the approximation to the value of \(e \approx 2.718281828\) with six terms? Does this infinite sum approach the value of \(e\) very quickly?

Find each logarithm. Give approximations to four decimal places. \(\ln 388.1\)
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