Question

Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: (a) \(0.00135 \mathrm{M} \mathrm{HNO}_{3}\), (b) \(0.425 \mathrm{~g}\) of \(\mathrm{HClO}_{4}\) in \(2.00 \mathrm{~L}\) of solution, \((\mathrm{c}) 5.00 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\) diluted to \(0.500 \mathrm{~L},(\mathrm{~d})\) a mixture formed by adding \(50.0 \mathrm{~mL}\) of \(0.020 \mathrm{M} \mathrm{HCl}\) to \(150 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{HI}\)

Short answer

The pH of the strong acid solutions are approximately: (a) 2.87 for the 0.00135M HNO3 solution. (b) 2.67 for the HClO4 solution with 0.425g in 2.00L. (c) 2.00 for the HCl solution after dilution (5.00mL of 1.00M HCl diluted to 0.500L). (d) 1.90 for the mixture formed by adding 50.0mL of 0.020M HCl to 150mL of 0.010M HI.

Step-by-step solution

Case (a): Finding the pH of 0.00135M HNO3 Solution

Since the concentration of HNO3 is given directly, we can use the formula to find the pH: \(\mathrm{pH} = -\log{[\mathrm{H}^+]} = -\log{(0.00135)} \approx 2.87\)

Case (b): Finding the pH of HClO4 Solution

First, convert the mass of HClO4 to moles: moles = \(\frac{\mathrm{mass}}{\mathrm{molar~mass}}\) moles = \(\frac{0.425}{100.46} \approx 0.00423\) Now, find the concentration of HClO4 in the solution: concentration = \(\frac{\mathrm{moles}}{\mathrm{volume(L)}} = \frac{0.00423}{2} = 0.00212 \mathrm{M}\) Calculate the pH of the solution: \(\mathrm{pH} = -\log{[\mathrm{H}^+]} = -\log{(0.00212)} \approx 2.67\)

Case (c): Finding the pH of HCl Solution After Dilution

Find the concentration of the HCl solution after dilution using the formula: \(C_1V_1 = C_2V_2\) \(C_2 = \frac{C_1V_1}{V_2} = \frac{(1.00)(0.005)}{(0.5)} \approx 0.010 \mathrm{M}\) Calculate the pH of the solution: \(\mathrm{pH} = -\log{[\mathrm{H}^+]} = -\log{(0.010)} = 2\)

Case (d): Finding the pH of the Mixture of HCl and HI Solutions

Calculate the moles of HCl: moles of HCl = (concentration of HCl)×(volume of HCl) = (0.020)(0.050) = 0.00100 Calculate the moles of HI: moles of HI = (concentration of HI)×(volume of HI) = (0.010)(0.150) = 0.00150 Find the total moles of hydrogen ions: total moles of hydrogen ions = moles of HCl + moles of HI = 0.00100 + 0.00150 = 0.00250 Find the total volume of the mixture: 50.0 mL of HCl + 150 mL of HI = 200 mL = 0.200 L Find the total concentration of hydrogen ions: \(\mathrm{[H^+]} = \frac{\mathrm{total~moles~of~hydrogen~ions}}{\mathrm{total~volume}} = \frac{0.00250}{0.200} \approx 0.0125 \mathrm{M}\) Calculate the pH of the solution: \(\mathrm{pH} = -\log{[\mathrm{H}^+]} = -\log{(0.0125)} \approx 1.90\) The pH of the solutions are approximately as follows: (a) 2.87 (b) 2.67 (c) 2.00 (d) 1.90

Key concepts

pH Level

Understanding the pH level is crucial for both chemistry students and professionals, as it determines the acidity or basicity of a solution. The pH scale ranges from 0 to 14, with 7 being neutral. Solutions with a pH less than 7 are acidic, while those with a pH greater than 7 are basic or alkaline.

For example, when you calculate the pH of a 0.00135 M HNO3 solution, which is a strong acid, you are essentially measuring the concentration of hydrogen ions (H+) in the solution. In this case, the calculation of pH as approximately 2.87 directly indicates the acidic nature of the solution.

Remember, the pH scale is logarithmic; this means each whole pH value below 7 is 10 times more acidic than the next higher value. Thus, a solution with a pH of 2 is ten times more acidic than one with a pH of 3. It's this logarithmic nature that makes the pH scale efficient in measuring a wide range of hydrogen ion concentrations.

Molarity and Dilution

Molarity, a term often used interchangeably with 'molar concentration', refers to the number of moles of a solute per liter of solution. It's represented by the symbol M and is a critical factor in the calculation of the pH of a solution. Dilution, on the other hand, involves adding more solvent to a solution, thus reducing its concentration. The concept of molarity and dilution is based on the principle that the number of moles of solute remains the same before and after the dilution process.

Take the dilution of 5.00 mL of 1.00 M HCl to 0.500 L as an example; the final molarity is calculated to be 0.010 M. Although the volume of the solution has increased, the number of moles of HCl remains constant, explaining why the formula (\(C_1V_1 = C_2V_2\)) is so pivotal. It essentially states that the product of the initial concentration (\(C_1\)) and volume (\(V_1\)) equals the final concentration (\(C_2\)) times the final volume (\(V_2\)). This relationship is fundamental when you're preparing solutions in a lab or analyzing the change in pH due to dilution.

Logarithmic pH Scale

The pH scale is inherently logarithmic, designed to quantify the acidity or basicity of solutions in a manageable way. Since the range of hydrogen ion concentrations in solutions can be vast, the logarithmic scale compresses these variations into a manageable range of numbers from 0 to 14.

This is shown explicitly in the formula used to calculate pH: (\(pH = -\log{[H^+]}\)). The negative sign in the formula is important as it makes the pH value a positive number. For example, a highly acidic solution with a high concentration of hydrogen ions (\( [H^+] \)) will result in a lower pH value, as seen with the \( pH \) calculated for the mixture of HCl and HI to be around 1.90. The logarithmic nature of the pH scale means that even small changes in pH represent significant changes in the concentration of hydrogen ions. This is why precision in pH calculations is so important in scientific contexts, be it for a simple classroom experiment or a critical industrial process.