Problem 6 - Analytical Chemistry

Question

Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of the following strong acid solutions: (a) \(0.020 \mathrm{M} \mathrm{HClO}_{4}\), (b) \(1.3 \times 10^{-4} \mathrm{M} \mathrm{HNO}_{3}\) (c) \(1.2 \mathrm{M} \mathrm{HCl}\). (d) \(1.2 \times 10^{-9} M \mathrm{HCl}\) (e) \(2.4 \times 10^{-7} \mathrm{MHNO}_{3}\)

Step-by-Step Solution

Answer

(a) pH=1.70, pOH=12.30; (b) pH=3.89, pOH=10.11; (c) pH=-0.08, pOH=14.08; (d) pH≈7, pOH≈7; (e) pH≈6.62, pOH≈7.38.

Step 1: Understanding pH and pOH

The pH of a solution is calculated using the formula: \( \text{pH} = -\log[\text{H}^+] \), where \( [\text{H}^+] \) is the concentration of hydrogen ions. The pOH is calculated as \( \text{pOH} = -\log[\text{OH}^-] \). For strong acids, we assume full dissociation in water, meaning the concentration of the acid is equal to the concentration of \( [\text{H}^+] \).

Step 2: Calculating pH for Part (a)

For the solution of \( 0.020 \, \text{M} \, \text{HClO}_4 \), the concentration of \( [\text{H}^+] = 0.020 \, \text{M} \). Thus, \( \text{pH} = -\log(0.020) = 1.70 \).

Step 3: Calculating pOH for Part (a)

Using the relation \( \text{pH} + \text{pOH} = 14 \), we find \( \text{pOH} = 14 - 1.70 = 12.30 \).

Step 4: Calculating pH for Part (b)

For \( 1.3 \times 10^{-4} \, \text{M} \, \text{HNO}_3 \), \( [\text{H}^+] = 1.3 \times 10^{-4} \). Therefore, \( \text{pH} = -\log(1.3 \times 10^{-4}) \approx 3.89 \).

Step 5: Calculating pOH for Part (b)

Using \( \text{pH} + \text{pOH} = 14 \), \( \text{pOH} = 14 - 3.89 \approx 10.11 \).

Step 6: Calculating pH for Part (c)

For \( 1.2 \, \text{M} \, \text{HCl} \), \( [\text{H}^+] = 1.2 \). Hence, \( \text{pH} = -\log(1.2) \approx -0.08 \).

Step 7: Calculating pOH for Part (c)

We use \( \text{pH} + \text{pOH} = 14 \) to find \( \text{pOH} = 14 - (-0.08) = 14.08 \).

Step 8: Special Consideration for Part (d)

For very dilute acids like \( 1.2 \times 10^{-9} \, \text{M} \, \text{HCl} \), water's own \( [\text{H}^+] \approx 10^{-7} \) affects the \( \text{pH} \). This causes the solution to have roughly neutral \( \text{pH} = 7 \).

Step 9: Calculating pOH for Part (d)

Because the system is neutral, \( \text{pH} + \text{pOH} = 14 \) gives \( \text{pOH} \approx 7 \).

Step 10: Special Consideration for Part (e)

For \( 2.4 \times 10^{-7} \, \text{M} \, \text{HNO}_3 \), since it's near \( 10^{-7} \), the resulting \( \text{pH} \) will be slightly below 7, i.e., approximately 6.62, from water’s own contribution.

Step 11: Calculating pOH for Part (e)

With \( \text{pH} \approx 6.62 \), we find \( \text{pOH} = 14 - 6.62 \approx 7.38 \).

Key Concepts

Strong Acids

Strong acids are special because they completely separate into ions when they dissolve in water. This means that if you have a strong acid, you can assume it's fully broken down into hydrogen ions (\(\text{H}^+\)) and the respective anions. This concept is crucial when working with pH calculations because the stronger the acid, the more straightforward the calculation becomes. Remember:

• Strong acids like \(\text{HCl}, \text{HNO}_3, \text{and } \text{HClO}_4\) dissociate completely in water.
• This complete dissociation makes it simple to find out how many hydrogen ions are in the solution, as they’re equal to the acid's concentration.

So, when you're doing pH calculations with strong acids, just use the concentration of the acid to find the concentration of hydrogen ions.

pOH

pOH is related to pH and provides another way to look at the acidity or basicity of a solution. While pH indicates the concentration of hydrogen ions, pOH tells us how many hydroxide ions (\(\text{OH}^-\)) are present. The relationship between them is:

• \(\text{pH} + \text{pOH} = 14\)
• This equation holds true at 25°C, which is room temperature.
• If you know the pH, calculating the pOH is simply subtracting the pH value from 14, and vice versa.

Even though strong acids don't directly give us hydroxide ions, understanding pOH is useful, especially in neutral or basic solutions.

Dilute Acid Solutions

When dealing with very dilute acid solutions, things get a bit tricky. If an acid is very dilute, its concentration might be close to or less than that of the hydrogen ions naturally present in water. For instance:

• Water naturally has a hydrogen ion concentration of \(10^{-7}M\).
• If an acid's concentration is significantly lower than this, its impact might not change the natural pH significantly.

For example, with very dilute \(1.2 \times 10^{-9} M \text{HCl}\), the pH is mainly influenced by water. In such cases, the pH is close to neutral (around 7), rather than being an acidic pH as expected from an acid.

Hydrogen Ion Concentration

Hydrogen ion concentration determines the pH of a solution. The relationship is calculated using the formula:\[\text{pH} = -\log[\text{H}^+]\]This formula allows you to convert the concentration directly into a pH value, which is a more intuitive measure of acidity. Key points include:

• The more hydrogen ions in a solution, the lower the pH, making the solution more acidic.
• For strong acids, since they dissociate fully, the concentration of hydrogen ions is the same as the concentration of the acid.

Understanding this relationship is vital for accurately determining how acidic or basic a solution is based on its hydrogen ion concentration.