Question

Calculate the pH of each of the following strong acid solutions: \((\mathbf{a}) 0.0178 \mathrm{M} \mathrm{HNO}_{3},(\mathbf{b}) 0.500 \mathrm{~g}\) of \(\mathrm{HClO}_{3}\) in \(5.00 \mathrm{~L}\) of solution, \((\mathbf{c}) 5.00 \mathrm{~mL}\) of \(2.00 \mathrm{M} \mathrm{HCl}\) diluted to \(0.500 \mathrm{~L}\), (d) a mixture formed by adding \(75.0 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{HCl}\) to \(125 \mathrm{~mL}\) of \(0.020 \mathrm{M} \mathrm{HBr}\).

Short answer

The pH values for the given strong acid solutions are as follows: a) \(pH = 1.75\) b) \(pH = 2.93\) c) \(pH = 1.70\) d) \(pH = 1.79\)

Step-by-step solution

Calculate [H+] for 0.0178 M HNO3 solution

For strong acids, [H+] = the molarity of the acid: [H+] = 0.0178 M #Solution a: Calculate pH#

Calculate pH

Using the formula pH = -log[H+]: pH = -log(0.0178) pH = 1.75 #Solution b: Calculate moles of HClO3 #

Calculate moles of HClO3

HClO3 has a molar mass of 84.46 g/mol. 0.500 g HClO3 * (1 mol / 84.46 g) = 0.00592 mol #Solution b: Calculate [H+] (concentration)#

Calculate [H+] (concentration)

[H+] is equal to the moles of HClO3 per liter: [H+] = 0.00592 mol / 5.00 L [H+] = 0.00118 M #Solution b: Calculate pH#

Calculate pH

Using the formula pH = -log[H+]: pH = -log(0.00118) pH = 2.93 #Solution c: Calculate new concentration of HCl after dilution#

Calculate new concentration of HCl after dilution

M1V1 = M2V2 \[M2 = \frac{(2.00 M)(0.00500 L)}{0.500 L}\] M2 = 0.0200 M #Solution c: Calculate [H+] (concentration)#

Calculate [H+] (concentration)

In strong acid, [H+] = concentration of acid [H+] = 0.0200 M #Solution c: Calculate pH#

Calculate pH

Using the formula pH = -log[H+]: pH = -log(0.0200) pH = 1.70 #Solution d: Calculate combined moles of HCl and HBr#

Calculate combined moles of HCl and HBr

Moles of HCl = (0.010 M)(0.075 L) = 0.00075 mol Moles of HBr = (0.020 M)(0.125 L) = 0.00250 mol Total moles = 0.00075 + 0.00250 = 0.00325 mol #Solution d: Calculate combined volume of solution#

Calculate combined volume of solution

Total volume = 75.0 mL + 125 mL = 200 mL = 0.200 L #Solution d: Calculate [H+] (concentration)#

Calculate [H+] (concentration)

[H+] = total moles / total volume [H+] = 0.00325 mol / 0.200 L [H+] = 0.01625 M #Solution d: Calculate pH#

Calculate pH

Using the formula pH = -log[H+]: pH = -log(0.01625) pH = 1.79 Final pH values: a) 1.75 b) 2.93 c) 1.70 d) 1.79

Key concepts

Strong Acids

Strong acids are those that completely ionize in water, meaning they release hydrogen ions ([H+]) fully into the solution. This complete dissociation causes the concentration of hydrogen ions to be equal to the molarity of the acid itself. Examples of strong acids include hydrochloric acid (HCl), nitric acid (HNO3), and hydrobromic acid (HBr).

The pH of a solution containing a strong acid is directly influenced by the concentration of these released hydrogen ions.

• For example, a solution with a molarity of 0.0178 M HNO3 will have a [H+] concentration of 0.0178 M.
• The pH can then be determined using the formula: \[pH = -\log [H^+]\]

Thus, understanding the nature of strong acids can greatly simplify pH calculations, as seen in various examples and exercises.

Acid Concentration

Acid concentration refers to the amount of acid present in a given volume of solution, often expressed in moles per liter (Molarity, M). This concentration directly impacts the pH of strong acid solutions, as pH is a logarithmic scale that measures hydrogen ion activity.

To determine the acid concentration:

• Calculate the number of moles of the acid present.
• Divide the moles by the volume of the solution in liters to find the molarity.

For instance, if you have 0.500 g of HClO3 with a molar mass of 84.46 g/mol in a 5.00 L solution:

• Moles of HClO3 = 0.500 g / 84.46 g/mol = 0.00592 mol
• [H+] = 0.00592 mol / 5.00 L = 0.00118 M
• The pH is calculated using \[pH = -\log [H^+]\]

This process shows how acid concentration is foundational in calculating the pH of strong acid solutions.

Molarity

Molarity is a measurement of the concentration of a solute in a solution, defined as the number of moles of solute per liter of solution. It's a vital concept in chemistry that helps us understand the amount of a substance present in a solution.

To calculate molarity:

• Determine the moles of the solute.
• After determining the volume in liters, use the formula \[M = \frac{moles}{liters} \]

For example, in calculating the molarity of a solution for pH determination, the initial concentration of HCl in a dilution must be adjusted:

• If you start with 2.00 M HCl and dilute it, use \[M_1V_1 = M_2V_2\] to find the new molarity after dilution.
• This principle helps in exercises such as calculating the concentration of H+ after dilution when determining the pH.

Mastery of molarity provides a solid foundation for tackling various chemical calculations including pH.

Dilution in Chemistry

Dilution is the process of decreasing the concentration of a solute in a solution, usually by adding more solvent. It is a common practice in chemistry when lower concentrations are needed. The key relationship for dilution calculations is \[M_1V_1 = M_2V_2\], where:

• M₁ and V₁ are the molarity and volume of the starting solution.
• M₂ and V₂ are the molarity and volume after dilution.

For example, if you have 5.00 mL of 2.00 M HCl and it is diluted to 0.500 L:

• You can find the new molarity by \[M_2 = \frac{(2.00 \: \mathrm{M})(0.00500 \: \mathrm{L})}{0.500 \: \mathrm{L}} = 0.0200 \: \mathrm{M}\]
• Knowing the new concentration helps in calculating the pH as \[pH = -\log [H^+]\]

Understanding dilution is crucial in performing accurate pH calculations, especially in complex mixtures or when altering solution strengths.