Question

Calculate the \(\mathrm{pH}\) of each of the following solutions: (a) \(2.8 \times 10^{-4} \mathrm{MBa}(\mathrm{OH})_{2}\) (b) \(5.2 \times 10^{-4} \mathrm{M} \mathrm{HNO}_{3}\)

Short answer

(a) pH ≈ 10.75; (b) pH ≈ 3.28

Step-by-step solution

Understanding the Problem

We are asked to calculate the pH of two solutions. One is a solution of barium hydroxide, Ba(OH)₂, which is a strong base, and the other is a solution of nitric acid, HNO₃, which is a strong acid. To calculate the pH, we first need to understand the dissociation of each in water.

Dissociation of Ba(OH)₂

Ba(OH)₂ is a strong base and dissociates completely in water: Ba(OH)₂ → Ba²⁺ + 2OH⁻. This means that for every mole of Ba(OH)₂, two moles of OH⁻ are produced. Given the concentration of Ba(OH)₂ as 2.8 × 10⁻⁴ M, the concentration of OH⁻ is 2 × 2.8 × 10⁻⁴ M = 5.6 × 10⁻⁴ M.

Calculate pOH from [OH⁻]

The pOH is calculated using the formula: \[ ext{pOH} = - ext{log}_{10}[ ext{OH}^-] \]Substituting the [OH⁻] from Step 1 gives \[ ext{pOH} = - ext{log}_{10}(5.6 imes 10^{-4}) \]which approximately equals 3.25.

Convert pOH to pH for Ba(OH)₂

The relationship between pH and pOH is given by:\[ ext{pH} + ext{pOH} = 14 \]Substituting the calculated pOH gives \[ ext{pH} = 14 - 3.25 = 10.75 \].

Dissociation of HNO₃

HNO₃ is a strong acid and dissociates completely in water: HNO₃ → H⁺ + NO₃⁻. The concentration of H⁺ ions will be equal to the concentration of HNO₃, which is 5.2 × 10⁻⁴ M.

Calculate pH from [H⁺]

The pH is calculated using the formula: \[ ext{pH} = - ext{log}_{10}[ ext{H}^+] \]Substituting the [H⁺] from Step 4 gives \[ ext{pH} = - ext{log}_{10}(5.2 imes 10^{-4}) \]which approximately equals 3.28.

Key concepts

Strong Acids and Bases

Strong acids and bases are substances that completely dissociate into their constituent ions in water. This means that when you dissolve these substances in water, they break down entirely, releasing their ions.

For example:

• A strong acid like nitric acid, \( \mathrm{HNO}_3 \), dissociates completely to form hydrogen ions \( \mathrm{H}^+ \) and nitrate ions \( \mathrm{NO}_3^- \). The high concentration of \( \mathrm{H}^+ \) ions results in a low pH, indicating acidity.
• Similarly, a strong base like barium hydroxide, \( \mathrm{Ba(OH)}_2 \), dissociates in water to produce barium ions \( \mathrm{Ba}^{2+} \) and hydroxide ions \( \mathrm{OH}^- \). This results in a high concentration of \( \mathrm{OH}^- \) ions, contributing to basicity and a higher pH.

Understanding the complete dissociation of these compounds helps us calculate their contribution to the solution's pH or pOH.

Dissociation in Water

Dissociation is the process by which a compound breaks into its ions when dissolved in water. This is a key concept in understanding how strong acids and bases affect pH.

For a strong base like \( \mathrm{Ba(OH)}_2 \), the dissociation can be represented as:
\[ \mathrm{Ba(OH)}_2 \rightarrow \mathrm{Ba}^{2+} + 2\mathrm{OH}^- \]
This equation shows that each mole of \( \mathrm{Ba(OH)}_2 \) produces two moles of \( \mathrm{OH}^- \) ions. The concentration of \( \mathrm{OH}^- \) ions can be calculated by multiplying the base's molarity by 2.

In contrast, for nitric acid \( \mathrm{HNO}_3 \), the dissociation is:
\[ \mathrm{HNO}_3 \rightarrow \mathrm{H}^+ + \mathrm{NO}_3^- \]
Each mole of \( \mathrm{HNO}_3 \) results in one mole of \( \mathrm{H}^+ \) ions. The concentration of \( \mathrm{H}^+ \) ions is equal to the concentration of \( \mathrm{HNO}_3 \) initially provided.

The principle of dissociation is fundamental in calculating the pH or pOH of a solution.

pOH Calculation

The concept of pOH is crucial when dealing with bases, as it helps us understand their alkalinity. pOH is calculated from the concentration of hydroxide ions, \( [\mathrm{OH}^-] \), in a solution.

To calculate pOH, we use the formula:
\[ \mathrm{pOH} = -\log_{10}[\mathrm{OH}^-] \]
In our example with \( \mathrm{Ba(OH)}_2 \), we found the \( [\mathrm{OH}^-] = 5.6 \times 10^{-4} \mathrm{M} \). By applying this formula, the pOH is approximately 3.25.

This pOH value can then be used to find the pH of the solution, utilizing the relation:
\[ \mathrm{pH} + \mathrm{pOH} = 14 \]

Knowing how to convert between pH and pOH is vital for understanding the characteristics of solutions containing bases.

Logarithms in Chemistry

Logarithms play a significant role in chemistry, particularly in calculating pH and pOH. These measurements use logarithms to handle the wide range of hydrogen and hydroxide ion concentrations in solutions.

The definitions of pH and pOH are logarithmic scales:

• pH: \( \mathrm{pH} = -\log_{10}[\mathrm{H}^+] \)
• pOH: \( \mathrm{pOH} = -\log_{10}[\mathrm{OH}^-] \)

These equations mean that a ten-fold change in hydrogen or hydroxide ion concentration will result in a unit change in pH or pOH.

This logarithmic approach compresses the vast range of possible ion concentrations into a more manageable scale. It allows for easy representation and comparison of acidity or alkalinity levels in different solutions.

Logarithms help us interpret the pH scale, where values range from 0 (very acidic) to 14 (very basic), with 7 being neutral. Understanding how logarithms work is foundational to mastering calculations involving pH and pOH, as it allows chemists to express small numbers of hydrogen and hydroxide ions efficiently.