Problem 44 - Chemistry The Central Science In Si Units

Question

Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: (a) \(0.0178 \mathrm{M} \mathrm{HNO}_{3},\) (b) \(0.500 \mathrm{~g}\) of \(\mathrm{HClO}_{3}\) in \(5.00 \mathrm{~L}\) of solution, \((\mathbf{c}) 5.00 \mathrm{~mL}\) of \(2.00 \mathrm{M} \mathrm{HCl}\) diluted to \(0.500 \mathrm{~L}\), (d) a mixture formed by adding \(75.0 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{HCl}\) to \(125 \mathrm{~mL}\) of \(0.020 \mathrm{M}\) HBr.

Step-by-Step Solution

Answer

(a) pH = 1.75, (b) pH = 2.93, (c) pH = 1.70, (d) pH = 1.79

Step 1: Calculate pH for 0.0178 M HNO3

HNO3 is a strong acid and dissociates completely in water, so the concentration of hydrogen ions [H+] is equal to the concentration of the acid. Therefore, \[[\text{H}^+] = 0.0178 \, \text{M}\]The \(\text{pH}\) is calculated using the formula:\[\text{pH} = -\log_{10}[\text{H}^+]\]Substitute the concentration value:\[\text{pH} = -\log_{10}(0.0178) \approx 1.75\]

Step 2: Calculate pH for HClO3 in 5.00 L solution

First, calculate the number of moles of HClO3:\[\text{Moles of } \mathrm{HClO}_3 = \frac{0.500 \, \text{g}}{84.46 \, \text{g/mol}} = 0.00592 \, \text{mol}\]Convert to concentration \([\text{H}^+]\):\[[\text{H}^+] = \frac{0.00592 \, \text{mol}}{5.00 \, \text{L}} = 0.001184 \, \text{M}\]Now, calculate the \(\text{pH}\):\[\text{pH} = -\log_{10}(0.001184) \approx 2.93\]

Step 3: Calculate pH for diluted HCl solution

First, find the initial moles of HCl:\[\text{Moles of HCl} = 2.00 \, \text{M} \times 0.005 \, \text{L} = 0.010 \, \text{mol}\]After dilution to 0.500 L, the concentration \([\text{H}^+]\) becomes:\[[\text{H}^+] = \frac{0.010 \, \text{mol}}{0.500 \, \text{L}} = 0.020 \, \text{M}\]Calculate the \(\text{pH}\):\[\text{pH} = -\log_{10}(0.020) \approx 1.70\]

Step 4: Calculate pH for a mixture of HCl and HBr

First, find the moles of each acid in the solution:- Moles of HCl: \[0.010 \, \text{M} \times 0.075 \, \text{L} = 0.00075 \, \text{mol}\]- Moles of HBr:\[0.020 \, \text{M} \times 0.125 \, \text{L} = 0.0025 \, \text{mol}\]Total moles of \(\text{H}^+\):\[0.00075 \, \text{mol} + 0.0025 \, \text{mol} = 0.00325 \, \text{mol}\]Total volume of the mixture is:\[0.075 \, \text{L} + 0.125 \, \text{L} = 0.200 \, \text{L}\]Concentration of \(\text{H}^+\):\[[\text{H}^+] = \frac{0.00325 \, \text{mol}}{0.200 \, \text{L}} = 0.01625 \, \text{M}\]Calculate the \(\text{pH}\):\[\text{pH} = -\log_{10}(0.01625) \approx 1.79\]

Key Concepts

Strong Acids

Strong acids are a critical topic in chemistry. These substances dissociate completely in aqueous solutions. This means they release all their hydrogen ions into the solution, making them excellent conductors of electricity.
The key feature of strong acids is their ability to release nearly 100% of their hydrogen ions when dissolved. Examples include hydrochloric acid ( HCl ), nitric acid ( HNO_3 ), and sulfuric acid ( H_2SO_4 ).
When working with strong acids, calculating the pH is relatively straightforward because the concentration of hydrogen ions ( [H^+] ) is directly equal to the concentration of the acid. For instance, if you have a 0.1 M solution of HCl , [H^+] is equal to 0.1 M .

Molar Concentration

Molar concentration, often called molarity, is defined as the number of moles of solute per liter of solution. This measure is crucial in chemistry because it helps us understand concentrations and how strong or weak a solution may be.
To calculate molarity, you take the moles of the solute and divide this by the volume of the solution in liters: \( ext{Molarity (M)} = rac{ ext{moles of solute}}{ ext{liters of solution}}\)
For example, to find the molarity of a solution where 0.5 moles of solute are dissolved in 2 liters of solution, you would calculate: \( ext{M} = rac{0.5}{2} = 0.25 ext{ M}\). This calculation is pivotal for pH determination, particularly for strong acids.

Molarity Calculations

Performing molarity calculations is a common exercise in chemistry, especially when working with solutions. Molarity calculations involve a few steps:

• Identify the solute mass or given molarity.
• Convert the mass to moles if needed using molar mass.
• Divide the moles by the volume of the solution in liters.

For example, to find the molarity of a solution with 0.500 g of HClO_3 in 5.00 L, you first calculate the moles: \( ext{moles} = rac{0.500}{84.46} \, ext{g/mol} = 0.00592 \, ext{mol}\).Then, determine the concentration: \( ext{M} = rac{0.00592}{5.00} = 0.001184 \, ext{M}\). This skill allows chemists to prepare solutions with precise concentrations.

Acid Dissociation

Acid dissociation is an essential concept for understanding how acids behave in solutions. For strong acids, dissociation occurs completely. This refers to the breaking apart of an acid molecule to release H^+ ions and its conjugate base.
In a strong acid solution, like that of HCl, the dissociation can be represented by the equation: \( ext{HCl} \rightarrow ext{H}^+ + ext{Cl}^-\). This process happens extensively, and for most strong acids, nearly every molecule dissociates completely in an aqueous solution.
This complete dissociation allows chemists to equate the initial molarity of the acid directly with the concentration of hydrogen ions in the solution, simplifying pH calculations considerably.