Question

Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: (a) \(0.0167 M \mathrm{HNO}_{3},(\mathbf{b}) 0.225 \mathrm{g}\) of \(\mathrm{HClO}_{3}\) in 2.00 \(\mathrm{L}\) of solution, \((\mathbf{c}) 15.00 \mathrm{mL}\) 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 \(\mathrm{MHCl}\) to 125 \(\mathrm{mL}\) of 0.010 \(\mathrm{M} \mathrm{HI} .\)

Short answer

The approximate pH values for the given strong acid solutions are: (a) 1.78, (b) 2.87, (c) 1.52, and (d) 1.89.

Step-by-step solution

(a) Calculate the concentration of H+ ions in 0.0167M HNO3 solution.

Since \(\mathrm{HNO}_{3}\) is a strong acid, it will be completely ionized in water, and the concentration of \(\mathrm{H}^{+}\) ions will be the same as the initial concentration of the strong acid, in this case 0.0167M.

(a) Calculate the pH of 0.0167M HNO3 solution.

Using the formula \(\mathrm{pH} = - \log(\mathrm{H}^{+})\), we get: \[ \mathrm{pH} = - \log(0.0167) \approx 1.78 \]

(b) Calculate the moles of HClO3.

First, we will convert 0.225 g of \(\mathrm{HClO}_3\) to moles using its molar mass, which is approximately 84.46 g/mol: \[ \text{moles of } \mathrm{HClO}_{3} = \frac{0.225\,\mathrm{g}}{84.46\,\mathrm{g/mol}} \approx 0.00267\,\mathrm{mol} \]

(b) Calculate the concentration of H+ ions in 2.00L HClO3 solution.

We will now calculate the concentration of \(\mathrm{HClO}_3\) in the 2.00 L solution: \[ \text{concentration} = \frac{0.00267\,\mathrm{mol}}{2.00\,\mathrm{L}} \approx 0.00134\,\mathrm{M} \]

(b) Calculate the pH of the 0.225 g HClO3 in 2.00 L of solution.

Now we will use the formula \(\mathrm{pH} = - \log(\mathrm{H}^{+})\) to find the pH of the solution: \[ \mathrm{pH} = - \log(0.00134) \approx 2.87 \]

(c) Calculate the concentration of H+ ions in 15.00 mL of 1.00 M HCl diluted to 0.5 L.

First, we calculate the moles of \(\mathrm{HCl}\) in 15.00 mL of 1.00 M solution using the formula moles = Molarity × Volume: \[ \text{moles of } \mathrm{HCl} = 1.00\,\mathrm{M} \times 0.015\,\mathrm{L} = 0.015\,\mathrm{mol} \] Now we will dilute the solution to 0.5 L and recalculate the concentration of \(\mathrm{HCl}\): \[ \text{new concentration} = \frac{0.015\,\mathrm{mol}}{0.5\,\mathrm{L}} \approx 0.030\,\mathrm{M} \]

(c) Calculate the pH of the 15.00 mL of 1.00 M HCl diluted to 0.5 L.

Now we will use the formula \(\mathrm{pH} = - \log(\mathrm{H}^{+})\): \[ \mathrm{pH} = - \log(0.030) \approx 1.52 \]

(d) Calculate the total moles of H+ ions in the mixture of HCl and HI.

First, we determine the moles of each acid in their individual volumes: \[ \text{moles of } \mathrm{HCl} = 0.020\,\mathrm{M} \times 0.050\,\mathrm{L} = 0.0010\,\mathrm{mol} \] \[ \text{moles of } \mathrm{HI} = 0.010\,\mathrm{M} \times 0.125\,\mathrm{L} = 0.00125\,\mathrm{mol} \] Now, we add the moles of each acid to find the total moles of \(\mathrm{H}^+\) ions: \[ \text{total moles} = 0.0010\,\mathrm{mol} + 0.00125\,\mathrm{mol} = 0.00225\,\mathrm{mol} \]

(d) Calculate the concentration of H+ ions in the mixture.

Now, we will calculate the combined volume of the mixture: \[ \text{total volume} = 0.050\,\mathrm{L} + 0.125\,\mathrm{L} = 0.175\,\mathrm{L} \] Next, we will find the new concentration of \(\mathrm{H}^{+}\) ions in the mixture: \[ \text{new concentration} = \frac{0.00225\,\mathrm{mol}}{0.175\,\mathrm{L}} \approx 0.0129\,\mathrm{M} \]

(d) Calculate the pH of the mixture formed by adding 50.0 mL of 0.020 M HCl to 125 mL of 0.010 M HI.

Finally, we will use the formula \(\mathrm{pH} = - \log(\mathrm{H}^{+})\): \[ \mathrm{pH} = - \log(0.0129) \approx 1.89 \] In conclusion, the pH values for the different strong acid solutions are approximately: (a) 1.78, (b) 2.87, (c) 1.52, and (d) 1.89.

Key concepts

Understanding Strong Acid Solutions

Understanding what constitutes a strong acid solution is crucial when it comes to pH calculation. Strong acids are molecules that completely dissociate into their constituent ions when dissolved in water. This means that if you have a solution of a strong acid, virtually all of the acid will exist as hydrogen ions \(H^+\) and their counterions in the solution.

For instance, hydrochloric acid \(HCl\), nitric acid \(HNO_3\), and perchloric acid \(HClO_3\) are all considered strong acids. Because of their full dissociation, the concentration of hydrogen ions in a solution of a strong acid will be equal to the molarity of the acid. Hence, pH calculation for strong acid solutions becomes a straightforward task of taking the negative logarithm of the hydrogen ion concentration, expressed as \(\text{pH} = -\log\text{[}H^+\text{]}\).

In practice, for example, if you have a 0.0167 M solution of \(HNO_3\), you can straightaway deduce that the concentration of hydrogen ions will also be 0.0167 M, as no \(HNO_3\) molecules remain undissociated. This full dissociation principle is key to simplifying and solving pH problems related to strong acid solutions.

Mastering Molarity in Solution Calculations

Molarity, often denoted by \(M\), is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. Understanding molarity is vital for performing pH calculations, as it directly relates to the amount of acid in a solution.

To calculate molarity, you use the formula:
\[ \text{Molarity} = \frac{\text{number of moles of solute}}{\text{volume of solution in liters}} \]

When given a problem, such as calculating the pH of a 0.225 g sample of \(HClO_3\) in 2.00 L of solution, you must first find the number of moles of \(HClO_3\). This step involves converting the given mass of the acid to moles using its molar mass. Only then can you find the molarity by dividing the moles of \(HClO_3\) by the volume of the solution in liters. Finally, using the molarity, which represents the concentration of \(H^+\) ions for a strong acid, you can calculate the pH of the solution.

Remember that molarity is also central when dealing with dilutions or mixtures, as seen in problems where a certain volume of an acid is diluted with water or mixed with another acid. The concentration changes, but the fundamental concept of molarity does not.

The Concept of Dilution and Its Impact on Solution Concentration

Dilution is the process of reducing the concentration of a solution by adding more solvent, typically water, without changing the number of moles of solute present. It's an important concept in chemistry, particularly when solutions need to be prepared to a desired molarity.

The dilution formula is a straightforward expression:
\[ \text{M}_1 \times \text{V}_1 = \text{M}_2 \times \text{V}_2 \]

where \(\text{M}_1\) and \(\text{V}_1\) are the molarity and volume of the initial solution, and \(\text{M}_2\) and \(\text{V}_2\) are the molarity and volume after dilution. This formula is based on the principle that the number of moles of solute remains constant before and after dilution.

In the example of diluting 15.00 mL of 1.00 M \(HCl\) to 0.500 L, we use the initial molarity and volume to find the number of moles, then divide this value by the new volume to find the final concentration (molarity) after dilution. Knowing this new concentration allows us to calculate the new pH. Dilution fundamentally affects pH by changing the concentration of hydrogen ions in the solution, which is why understanding the dilution concept is crucial for accurate pH calculations in chemistry.