Question

To a \(10 \mathrm{~mL}\) of \(10^{-3} \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution water has been added to make the total volume of one litre. Its pOH would be : (a) 3 (b) 12 (c) 9 (d) 5

Short answer

The pOH of the diluted sulfuric acid solution would be approximately 9.

Step-by-step solution

Calculate the dilution of the sulfuric acid solution

After dilution, the concentration of the sulfuric acid solution is reduced. The initial concentration is given as a normality of \(10^{-3} \mathrm{N}\) for a volume of \(10 \mathrm{mL}\). After adding water to make up one liter (or \(1000 \mathrm{mL}\)), we can calculate the new concentration (C2) using the dilution formula \(C1V1 = C2V2\), where \(C1\) and \(V1\) are the initial concentration and volume, and \(C2\) and \(V2\) are the final concentration and volume, respectively.

Find the final concentration of the sulfuric acid solution

Using the dilution formula: \[(10^{-3} \mathrm{N}) * (10 \mathrm{mL}) = C2 * (1000 \mathrm{mL})\] Solving for \(C2\), we find that \(C2 = \frac{(10^{-3}) * 10}{1000} = 10^{-5} \mathrm{N}\). Since \(\mathrm{H}_{2}\mathrm{SO}_{4}\) is a strong diprotic acid, it donates two protons per molecule. Therefore, the molar concentration of \(\mathrm{H}^+\) ions is twice that of the normality, which is \(2 * 10^{-5} \mathrm{M}\).

Calculate the pH of the solution

The concentration of \(\mathrm{H}^+\) ions in the solution is \(2 * 10^{-5} \mathrm{M}\). To find the pH, use the formula \(\text{pH} = -\log[\mathrm{H}^+]\). Thus, \(\text{pH} = -\log(2 * 10^{-5})\).

Calculate the pOH of the solution

Knowing the relationship between pH and pOH, which is \(\text{pOH} = 14 - \text{pH}\), we can find the pOH. First calculate the pH using the given hydrogen ion concentration, then subtract that value from 14 to get the pOH.

Find the correct answer from the options

Using a calculator, we find the pH to be approximately \(\text{pH} = 4.7\). Therefore, \(\text{pOH} = 14 - 4.7 = 9.3\), which we round to \(9\) since the options are whole numbers. The closest option to our calculated pOH is (c) 9.

Key concepts

Dilution Formula in Chemistry

Understanding dilution is crucial when working with chemical solutions. Dilution refers to the process of reducing the concentration of a solute in a solution, usually by adding more solvent. The dilution formula, which is fundamental in chemistry, is given by the equation:
\[ C1V1 = C2V2 \]
In this formula, \( C1 \) and \( V1 \) are the initial concentration and volume of the solution, respectively, whereas \( C2 \) and \( V2 \) are the final concentration and volume after dilution.

To put this into context with our exercise, when water was added to the sulfuric acid solution to reach a total volume of one litre, we used the dilution formula to calculate the new concentration of the acid. This process is crucial in various fields such as pharmacy, where medication dosage needs precise dilutions, or in environmental science, where pollutant concentrations in water are measured and adjusted.

pH and pOH Relationship

The pH scale is a measure of the acidity or basicity of an aqueous solution. It can be calculated using the concentration of hydrogen ions \( [H^+] \) with the formula:
\[ \text{pH} = -\text{log}([H^+]) \]
Similarly, pOH measures the concentration of hydroxide ions and is determined by:
\[ \text{pOH} = -\text{log}([OH^-]) \]
There is a straightforward relationship between pH and pOH. In any aqueous solution at 25°C, their sum is always equal to 14:
\[ \text{pOH} = 14 - \text{pH} \]
This relation is crucial because once you know the concentration of hydrogen ions and thus the pH, you can easily find the pOH, and vice versa. Comprehending this relationship helps explain why the pOH of the diluted sulfuric acid solution was determined by subtracting the pH from 14 in our exercise.

Sulfuric Acid Properties

Sulfuric acid (\( H_2SO_4 \)) is known for its properties as a strong diprotic acid. Being diprotic means it can donate two protons (hydrogen ions) per molecule when it dissociates in water. As a result, the dissociation of sulfuric acid in water can be represented as follows:
\[ H_2SO_4 \rightarrow 2H^+ + SO_4^{2-} \]
This characteristic was utilized in the textbook exercise when the concentration of \( H^+ \) ions needed to be calculated. Since each molecule of sulfuric acid gives off two hydrogen ions, the molar concentration of \( H^+ \) is twice the normality of the sulfuric acid solution. This property is also essential when considering the acid's corrosiveness, reactivity, and its role in acid-base titrations. In industry, sulfuric acid is used in the manufacture of fertilizers, oil refining, and wastewater processing, demonstrating its importance and versatility.

Logarithm Applications in Chemistry

Logarithms offer a practical way to deal with the large range of values that we encounter in chemistry. This mathematical tool is especially useful when working with concepts like pH, which involves the negative logarithm of the hydrogen ion concentration.
\[ \text{pH} = -\text{log}([H^+]) \]
Logarithms help simplify calculations that would otherwise involve very small or large numbers, hence their frequent application in computations of pH, pOH, and in reaction kinetics. They allow chemists to move from exponential scales to manageable numbers, making it easier to compare the acidity or basicity of solutions, just like we saw with the pH calculation of the sulfuric acid solution in our textbook exercise. Understanding logarithms is therefore vital for students and professionals in the field of chemistry.