Problem 44 - Chemistry: The Central Science

Question

Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: (a) \(0.0167 \mathrm{MHNO}_{3}\), (b) \(0.225 \mathrm{~g}\) of \(\mathrm{HClO}_{3}\) in \(2.00 \mathrm{~L}\) of solution, (c) \(15.00 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\) diluted to \(0.500 \mathrm{~L}\), (d) a mixture formed by adding \(50.0 \mathrm{~mL}\) of \(0.020 \mathrm{M} \mathrm{HCl}\) to \(125 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{HI}\).

Step-by-Step Solution

Answer

The pH values for each strong acid solution are as follows: (a) \(pH \approx 1.78\) (b) \(pH \approx 2.88\) (c) \(pH \approx 1.52\) (d) \(pH \approx 1.89\)

Step 1: (a) Calculate the concentration of HNO3 solution

Here, we're given that the concentration of \(\mathrm{HNO}_{3}\) is \(0.0167\mathrm{M}\). Since strong acids completely ionize in water, we can conclude that: \[[\mathrm{H^+}] = 0.0167~\mathrm{M}\]

Step 2: (a) Calculate the pH of 0.0167M HNO3 solution

We can now use the pH formula: \[pH = -\log_{10} [\mathrm{H^+}] = -\log_{10} (0.0167) \] After calculating, we get: \[pH \approx 1.78\]

Step 3: (b) Calculate the concentration of HClO3 solution

To find the concentration of \(\mathrm{HClO}_{3}\), we need to convert the given mass of the acid to moles, and then divide by the volume of the solution. First, we find the molar mass of \(\mathrm{HClO}_{3}\): \[M_{\mathrm{HClO}_{3}} = 1.01 (\mathrm{H}) + 35.45 (\mathrm{Cl}) + 3 \times 16.00 (\mathrm{O}) \approx 84.46~\mathrm{g/mol}\] Now we can calculate the moles of \(\mathrm{HClO}_{3}\): \[\text{moles} = \frac{0.225 \mathrm{~g}}{84.46 \mathrm{~g/mol}} \approx 0.00266 \mathrm{~mol}\] Finally, we find the concentration by dividing the moles by the volume of the solution: \[[\mathrm{HClO}_{3}] = \frac{0.00266}{2.00 \mathrm{~L}} \approx 0.00133~\mathrm{M}\]

Step 4: (b) Calculate the pH of the HClO3 solution

Since \(\mathrm{HClO}_{3}\) is a strong acid, we have: \[[\mathrm{H^+}] = 0.00133~\mathrm{M}\] Now we can find the pH using the formula: \[pH = -\log_{10} [\mathrm{H^+}] = -\log_{10} (0.00133) \] After calculating, we get: \[pH \approx 2.88\]

Step 5: (c) Calculate the concentration of the HCl solution after dilution

We are given the initial concentration and volume of the \(\mathrm{HCl}\) solution before dilution, as well as the final volume after dilution. To calculate the final concentration, we can use the dilution formula: \[C_{1}V_{1} = C_{2}V_{2}\] Plug in the given values: \[(1.00\mathrm{M})(0.015 \mathrm{~L}) = C_{2}(0.500 \mathrm{~L})\] Now, solve for \(C_{2}\) (the concentration of \(\mathrm{HCl}\) after dilution): \[C_{2} = \frac{(1.00\mathrm{M})(0.015 \mathrm{~L})}{0.500 \mathrm{~L}} = \frac{0.015}{0.500} \approx 0.030\mathrm{M}\]

Step 6: (c) Calculate the pH of the diluted HCl solution

Since \(\mathrm{HCl}\) is a strong acid, we have: \[[\mathrm{H^+}] = 0.030\mathrm{M}\] Now we can find the pH using the formula: \[pH = -\log_{10} [\mathrm{H^+}] = -\log_{10} (0.030) \] After calculating, we get: \[pH \approx 1.52\]

Step 7: (d) Calculate the concentration of the mixture of HCl and HI solutions

To find the combined concentration of the two strong acid solutions, we can calculate the moles of both \(\mathrm{HCl}\) and \(\mathrm{HI}\), then divide by the total volume of the mixture: Moles of \(\mathrm{HCl}\): \[\text{moles}_{\mathrm{HCl}} = (0.020 \mathrm{M})(0.050 \mathrm{~L}) = 0.0010 \mathrm{~mol}\] Moles of \(\mathrm{HI}\): \[\text{moles}_{\mathrm{HI}} = (0.010 \mathrm{M})(0.125 \mathrm{~L}) = 0.00125 \mathrm{~mol}\] Total moles and volume of the mixture: \[\text{Total Moles} = 0.0010 + 0.00125 = 0.00225 \mathrm{~mol};\] \[\text{Total Volume} = 0.050 \mathrm{~L} + 0.125 \mathrm{~L} = 0.175 \mathrm{~L}\] Combined acid concentration: \[[\text{Acids}] = \frac{0.00225 \mathrm{~mol}}{0.175 \mathrm{~L}} \approx 0.0129\mathrm{M}\]

Step 8: (d) Calculate the pH of the mixture of HCl and HI solutions

Since both \(\mathrm{HCl}\) and \(\mathrm{HI}\) are strong acids, we have: \[[\mathrm{H^+}] = 0.0129\mathrm{M}\] Now we can find the pH using the formula: \[pH = -\log_{10} [\mathrm{H^+}] = -\log_{10} (0.0129) \] After calculating, we get: \[pH \approx 1.89\] To summarize, the pH values for each solution are: (a) 1.78 (b) 2.88 (c) 1.52 (d) 1.89

Key Concepts

pH Scale

The pH scale is a measure of the acidity or basicity of a solution. It runs from 0 to 14, with 7 being neutral. Solutions with a pH less than 7 are acidic, while those greater than 7 are basic. The scale is logarithmic, meaning each whole pH value below 7 is ten times more acidic than the next higher value. For instance, a solution with a pH of 3 is ten times more acidic than one with a pH of 4.

To calculate the pH of a solution, you use the formula:
\[\begin{equation} pH = -\text{log}_{10} [\text{H}^+] \end{equation}\] Here, \[\begin{equation} [\text{H}^+] \end{equation}\] represents the concentration of hydrogen ions in moles per liter (M). For strong acids, the \[\begin{equation} [\text{H}^+] \end{equation}\] value is equal to the concentration of the acid because they completely ionize in water.

Strong Acid Ionization

Strong acids are characterized by their ability to completely ionize in solution. This means that every molecule of a strong acid dissociates to form hydrogen ions (\[\begin{equation} \text{H}^+ \end{equation}\] ) and its corresponding anions. For example, hydrochloric acid (\[\begin{equation} \text{HCl} \end{equation}\] ) in water dissociates completely to form \[\begin{equation} \text{H}^+ \end{equation}\] and \[\begin{equation} \text{Cl}^- \end{equation}\] ions. Understanding this concept is vital when calculating the pH of strong acid solutions because it simplifies the process, allowing us to assume that the concentration of the acid is equal to the concentration of \[\begin{equation} \text{H}^+ \end{equation}\] ions directly.

Molar Mass Calculation

To calculate the molar concentration of a solution when given the mass of a solute, you must first determine the molar mass of the solute. The molar mass is the weight of one mole of a substance and is expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in a molecule of the substance. For instance, for \[\begin{equation} \text{HClO}_3 \end{equation}\] , the molar mass would be the sum of the atomic masses of hydrogen (H), chlorine (Cl), and three oxygen (O) atoms.

Once you've found the molar mass, you convert the mass of the solute to moles by dividing it by the molar mass. This step is crucial in the process of finding the molarity (moles per liter) of a solution and subsequently calculating the pH.

Acid Dilution

Dilution is the process of reducing the concentration of a solute in a solution, usually by adding more solvent. The relationship between the initial and final concentrations and volumes in a dilution can be expressed by the formula:
\[\begin{equation} C_1V_1 = C_2V_2 \end{equation}\] where \[\begin{equation} C_1 \end{equation}\] and \[\begin{equation} C_2 \end{equation}\] are the initial and final concentrations, and \[\begin{equation} V_1 \end{equation}\] and \[\begin{equation} V_2 \end{equation}\] are the initial and final volumes, respectively. This equation is crucial for accurately determining the concentration of an acid after dilution, an essential step before calculating the pH of a diluted acid solution.