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 \(0.500 \mathrm{~L}\), (d) a mixture formed by adding \(50.0 \mathrm{~mL}\) of \(0.020 \mathrm{M} \mathrm{HCl}\) to \(150 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{HL}\)

Short answer

The calculated pH values for each acid solution are as follows: (a) pH = 2.87 for the 0.00135 M HNO3 solution (b) pH = 2.67 for the 0.425 g HClO4 in 2.00 L solution (c) pH = 2.00 for the 5.00 mL of 1.00 M HCl diluted to 0.500 L (d) pH = 1.90 for the mixture of 50.0 mL of 0.020 M HCl and 150 mL of 0.010 M HL.

Step-by-step solution

(a) Find the concentration of HNO3

The concentration of HNO3 is given as \(0.00135 \mathrm{M}\). Since it is a strong acid, it dissociates completely into ions. Therefore, the concentration of hydrogen ions \([\mathrm{H^+}]\) is the same as the concentration of the acid, \(0.00135 \mathrm{M}\).

(a) Calculate the pH of HNO3 solution

Now we can use the pH formula: pH = -log(\([\mathrm{H^+}]\)) pH = -log(\(0.00135\)) pH = 2.87

(b) Calculate the moles of HClO4

First, we need to find the moles of HClO4 from the given mass and its molar mass. Moles = \(\frac{mass}{molar mass}\) = \(\frac{0.425 \mathrm{~g}}{100.46 \mathrm{~g/mol}}=0.00423 \mathrm{mol}\)

(b) Determine the concentration of HClO4

Now we can find the concentration of HClO4 in the 2.00 L solution: Concentration = \(\frac{moles}{volume}\) = \(\frac{0.00423 \mathrm{~mol}}{2.00 \mathrm{~L}} = 0.00212 \mathrm{M}\) Since HClO4 is a strong acid, \([\mathrm{H^+}]\) = 0.00212 M.

(b) Calculate the pH of HClO4 solution

Now we can use the pH formula: pH = -log(\(0.00212\)) pH = 2.67

(c) Calculate the concentration of HCl before dilution

The initial concentration of HCl is given as \(1.00 \mathrm{M}\) and the volume is \(5.00 \mathrm{ mL}\) (0.005 L).

(c) Determine the moles of HCl

Find the moles of HCl in the initial solution: Moles = Concentration × Volume = \((1.00 \mathrm{M}) (0.005 \mathrm{L}) = 0.005 \mathrm{mol}\)

(c) Calculate the concentration after dilution

Determine the concentration of HCl after dilution to 0.500 L: Concentration = \(\frac{moles}{volume}\) = \(\frac{0.005 \mathrm{mol}}{0.500 \mathrm{L}} = 0.010 \mathrm{M}\) Since HCl is a strong acid, \([\mathrm{H^+}]\) = 0.010 M.

(c) Calculate the pH of HCl solution after dilution

Now we can use the pH formula: pH = -log(\(0.010\)) pH = 2.00

(d) Calculate the moles of HCl and HL

First, we need to calculate the moles of HCl and HL in the initial solution volumes: Moles of HCl = Concentration × Volume = \((0.020 \mathrm{M}) (0.050 \mathrm{L}) = 0.0010 \mathrm{mol}\) Moles of HL = Concentration × Volume = \((0.010 \mathrm{M}) (0.150 \mathrm{L}) = 0.0015 \mathrm{mol}\)

(d) Calculate the total volume of the mixture

The total volume of the mixture is the sum of the initial volumes: \(0.050 \mathrm{L} + 0.150 \mathrm{L} = 0.200 \mathrm{L}\)

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

Since both HCl and HL are strong acids, they both dissociate completely in the solution. Thus, the total moles of H+ ions in the mixture are the sum of the moles of the two acids: \(0.0010 \mathrm{mol} + 0.0015 \mathrm{mol} = 0.0025 \mathrm{mol}\). To find the concentration of H+ ions in the mixture, divide the total moles of H+ ions by the total volume: \([\mathrm{H^+}]\) = \(\frac{0.0025 \mathrm{mol}}{0.200 \mathrm{L}} = 0.0125 \mathrm{M}\)

(d) Calculate the pH of the mixture

Now we can use the pH formula: pH = -log(\(0.0125\)) pH = 1.90

Key concepts

Strong Acid Solutions

When discussing strong acid solutions, we're talking about acids that completely dissociate into their ions in water. This complete dissociation implies that the hydrogen ion concentration, \( [H^+] \), in the solution is equal to the original concentration of the acid.

For example, if you have a solution of hydrochloric acid (HCl) with a molar concentration of 1 M, this means that every molecule of HCl will separate into one hydrogen ion (H+) and one chloride ion (Cl-), resulting in a \( [H^+] \), of 1 M as well. The implication for pH calculations is straightforward; the molar concentration of the acid directly gives you the molar concentration of hydrogen ions necessary to calculate pH.

To measure how acidic the solution is, we calculate the pH, which is the negative logarithm of the hydrogen ion concentration: \( pH = -\log([H^+]) \). The lower the pH value, the more acidic the solution is. Strong acids like sulfuric acid (\( H_2SO_4 \)), nitric acid (\( HNO_3 \)), and hydrochloric acid (\( HCl \) are all well-known examples that fall into this category.

Molar Concentration

Molar concentration, also known as molarity, is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute divided by the volume of the solution in liters (L). Mathematically, it is expressed as: \( C = \frac{n}{V} \), where \( C \) is the concentration in moles per liter (M), \( n \) is the number of moles, and \( V \) is the volume in liters.

In the context of our exercise, to calculate the molar concentration of an acid like HClO4, you'd need to know the mass of the acid used and its molar mass. Once you have the number of moles of the acid, you can divide by the solution's volume to find its molarity.

For instance, if you dissolve a certain mass of an acid in a known volume of water, you first convert the mass to moles (using the molar mass) and then divide by the volume to get molarity. This value becomes crucial when calculating the pH of the solution, as seen in steps concerning the dilution of HCl. Dilution, which involves adding more solvent to a solution, decreases the molarity of the solution, and consequently the \( [H^+] \) and the pH.

Acid Dissociation

The term acid dissociation refers to the process by which an acid releases hydrogen ions (\( H^+ \) into a solution. In the case of strong acids, this is an all-or-nothing affair because they completely dissociate in water. That means each molecule of the strong acid will contribute one (or more, depending on the acid) \( H^+ \) ion to the solution.

Understanding this process is critical when performing pH calculations. We assume that the molar concentration of hydrogen ions is equal to that of the original strong acid for strong acids. For instance, in our exercise-related problems, when calculating the final concentration of \( H^+ \) ions in mixtures or after dilution, we add up the moles of \( H^+ \) provided by each acid present, given that they fully dissociate, and then we use this concentration in the pH calculation.

The acid dissociation concept is essential for solving problems involving mixtures of acids or serial dilutions. It helps us appreciate that, regardless of the amount of solvent, when we're dealing with strong acids, the number of moles of \( H^+ \) will solely depend on the initial moles of acid we started with, as every single one of those moles will dissociate fully.