Problem 43 - Chemistry The Central Science

Question

Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: \((\mathbf{a}) 8.5 \times 10^{-3} \mathrm{M} \mathrm{HBr},(\mathbf{b}) 1.52 \mathrm{g}\) of \(\mathrm{HNO}_{3}\) in 575 \(\mathrm{mL}\) of solution, \((\mathbf{c}) 5.00 \mathrm{mL}\) of 0.250 \(\mathrm{M} \mathrm{ClO}_{4}\) diluted to 50.0 \(\mathrm{mL}\) (d) a solution formed by mixing 10.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HBr}\) with 20.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{HCl} .\)

Step-by-Step Solution

Answer

The pH values for the strong acid solutions are as follows: (a) \(\mathrm{pH} = 2.07\) (b) \(\mathrm{pH} = 1.38\) (c) \(\mathrm{pH} = 1.60\) (d) \(\mathrm{pH} = 0.78\)

Step 1: Calculate the concentration of \(\mathrm{H^{+}}\) ions in solution

HBr is a strong acid. Therefore, it will dissociate completely into \(\mathrm{H^{+} (aq)}\) and \(\mathrm{Br^{-} (aq)}\) ions in solution. The concentration of \(\mathrm{H^{+}}\) ions is the same as the concentration of HBr, which is \(8.5 \times 10^{-3} \mathrm{M}\).

Step 2: Calculate the pH

Use the formula \(\mathrm{pH} = -\log(\mathrm{[H^{+}]})\) to calculate the pH of the solution. In this case, \(\mathrm{pH} = -\log(8.5 \times 10^{-3})\). Calculate the pH: \[\mathrm{pH} = 2.07\] #b) Calculate the pH of 1.52 g of HNO₃ in 575 mL of solution:

Step 1: Calculate the concentration of \(\mathrm{H^{+}}\) ions in solution

First, find the moles of HNO₃ by dividing the mass by the molar mass (63.01 g/mol): \(\frac{1.52\,\mathrm{g}}{63.01\, \mathrm{g/mol}} = 0.0241\,\mathrm{mol}\) Next, convert the volume of solution to liters: \(\frac{575\,\mathrm{mL}}{1000} = 0.575\,\mathrm{L}\) Now, calculate the molar concentration of HNO₃: \(\frac{0.0241\,\mathrm{mol}}{0.575\,\mathrm{L}} = 0.0419\,\mathrm{M}\) Since HNO₃ is a strong acid, the concentration of \(\mathrm{H^{+}}\) ions in the solution is the same as the concentration of HNO₃, which is \(0.0419\,\mathrm{M}\).

Step 2: Calculate the pH

Use the formula \(\mathrm{pH} = -\log(\mathrm{[H^{+}]})\) to calculate the pH of the solution. In this case, \(\mathrm{pH} = -\log(0.0419)\). Calculate the pH: \[\mathrm{pH} = 1.38\] #c) Calculate the pH of 5.00 mL of 0.250 M ClO₄⁻ diluted to 50.0 mL:

Step 1: Calculate the concentration of \(\mathrm{H^{+}}\) ions in solution

First, find the moles of HClO₄ in 5.00 mL of 0.250 M solution: \((0.250\,\mathrm{M})(0.00500\,\mathrm{L}) = 0.00125\,\mathrm{mol}\) Next, calculate the new concentration after dilution: \(\frac{0.00125\,\mathrm{mol}}{0.0500 \,\mathrm{L}} = 0.0250\,\mathrm{M}\) Since HClO₄ is a strong acid, the concentration of \(\mathrm{H^{+}}\) ions is the same as the concentration of HClO₄, which is \(0.0250\,\mathrm{M}\).

Step 2: Calculate the pH

Use the formula \(\mathrm{pH} = -\log(\mathrm{[H^{+}]})\) to calculate the pH of the solution. In this case, \(\mathrm{pH} = -\log(0.0250)\). Calculate the pH: \[\mathrm{pH} = 1.60\] #d) Calculate the pH of a solution formed by mixing 10.0 mL of 0.100 M HBr with 20.0 mL of 0.200 M HCl:

Step 1: Calculate the concentration of \(\mathrm{H^{+}}\) ions in solution

First, calculate the moles of \(\mathrm{H^+}\) ions from HBr and HCl: HBr: \((0.100\,\mathrm{M})(0.0100\,\mathrm{L}) = 0.00100\,\mathrm{mol}\) HCl: \((0.200\,\mathrm{M})(0.0200\,\mathrm{L}) = 0.00400\,\mathrm{mol}\) Next, add the moles of \(\mathrm{H^+}\) ions: \(0.00100\,\mathrm{mol} + 0.00400\,\mathrm{mol} = 0.00500\,\mathrm{mol}\) Now, calculate the total volume of the solution: \(10.0\,\mathrm{mL} + 20.0\,\mathrm{mL} = 30.0\,\mathrm{mL} = 0.0300\,\mathrm{L}\) Calculate the new concentration of \(\mathrm{H^+}\) ions: \(\frac{0.00500\,\mathrm{mol}}{0.0300\,\mathrm{L}} = 0.167\,\mathrm{M}\)

Step 2: Calculate the pH

Use the formula \(\mathrm{pH} = -\log(\mathrm{[H^{+}]})\) to calculate the pH of the solution. In this case, \(\mathrm{pH} = -\log(0.167)\). Calculate the pH: \[\mathrm{pH} = 0.78\]

Key Concepts

pH Calculation

The concept of pH is a numerical scale that represents the acidity or basicity of a solution. The pH formula is defined as the negative logarithm to the base 10 of the concentration of hydrogen ions (\([H^+]\)). In simpler terms, it measures how acidic or basic a solution is. When handling strong acids like HBr, HNO₃, or HClO₄, they completely dissociate in water.For instance:

• To find the pH of a solution where the concentration of hydrogen ions is known, use the formula \( \text{pH} = -\log{([H^+])} \).
• A lower pH value implies a stronger acid and a higher concentration of hydrogen ions.
• For strong acids, the pH is typically less than 7, often ranging from 0 to about 3 for concentrated solutions.

By using these simple mathematical steps and understanding the correlation between hydrogen ion concentration and pH, one can successfully determine the acidity of any strong acid solution.

Molar Concentration

Molar concentration, or molarity, is the measure of the concentration of a solute in a solution. It is denoted as moles of solute per liter of solution, represented as M (mol/L). In the context of strong acids, knowing the molarity is crucial because it directly relates to the hydrogen ion concentration, which helps in pH calculation.To calculate molarity:

• Determine the number of moles of solute, which for acids typically involves the acid dissociating to form hydrogen ions.
• Convert the solution's volume to liters, and divide the moles of solute by this volume.

For example, if you have 1.52 grams of HNO₃ dissolved in 575 mL of water, convert the mass to moles using the molar mass, find the solution volume in liters, and use the formula \( \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \) to find the molarity.

Dilution

Dilution is a common laboratory procedure used to decrease the concentration of a solution by adding more solvent. When diluting a strong acid, the number of moles of solute remains constant, but the total volume of the solution increases, thus lowering the concentration.The dilution formula \( C_1V_1 = C_2V_2 \) helps to calculate new concentration:

• \( C_1 \) and \( V_1 \) are the initial concentration and volume.
• \( C_2 \) and \( V_2 \) are the final concentration and volume after dilution.

For example, starting with 0.250 M HClO₄ and diluting from 5.00 mL to 50.0 mL, use this formula to find the new concentration (which is also the new hydrogen ion concentration because it’s a strong acid). This demonstrates how dilution affects molarity and pH without altering the amount of acid present.

Acid Dissociation

Acid dissociation refers to the process by which an acid dissolves in water and releases hydrogen ions (\([H^+]\)). Strong acids completely dissociate in solutions, meaning each acid molecule breaks down to release its hydrogen ions.Here's how acid dissociation impacts calculations:

• With strong acids like HBr and HCl, the original concentration of the acid is the same as the concentration of hydrogen ions released.
• This simplifies pH calculations immensely as it circumvents the need for an equilibrium constant (K_a).
• In mixtures, such as combining different strong acids, you sum the moles of hydrogen ions before calculating total concentration for pH.

Understanding dissociation is crucial because it directly influences how we calculate the acidity of solutions, ensuring accurate pH values and hence providing a clear picture of the solution's acidity level.