Problem 26
Question
Find \(\left[\mathrm{H}^{+}\right]\) and the \(\mathrm{pH}\) of the following solutions. (a) \(1.75 \mathrm{~L}\) of a \(37.5 \%\) (by mass) solution \((d=1.00 \mathrm{~g} / \mathrm{mL})\) of \(\mathrm{HCl} .\) What is the \(\mathrm{pH}\) of \(0.175 \mathrm{~L}\) of the same solution? (b) A solution made up of \(22 \mathrm{~g}\) of \(\mathrm{HBr}\) dissolved in enough water to make \(479 \mathrm{~mL}\) of solution. What is the \(\mathrm{pH}\) if the same mass of \(\mathrm{HBr}\) is dissolved in enough water to make \(47.9 \mathrm{~mL}\) of solution?
Step-by-Step Solution
Verified Answer
Question: Calculate the concentration of H+ ions and the pH of the following solutions: (a) a 1.75 L solution with a density of 1.00 g/mL and a 37.5% by mass composition of HCl, and (b) 22 g of HBr dissolved in 479 mL of water. Additionally, find the pH of 0.175 L of the same HCl solution from part (a).
Answer: For part (a), the concentration of H+ ions is 10.3 M, and the pH is approximately -1.01 for both the 1.75 L solution and the 0.175 L solution. For part (b), the concentration of H+ ions is 0.568 M, and the pH is approximately 0.25.
1Step 1: Calculate the molarity of the acid in each solution
For part (a), we have a 1.75 L solution with a density of 1.00 g/mL and a 37.5% by mass composition of HCl. First, convert the volume into mass:
Mass of the solution = volume x density
Mass of the solution = (1.75 L x 1000 mL/L) x 1.00 g/mL = 1750 g
Now find the mass of HCl:
Mass of HCl = (0.375)(1750 g) = 656.25 g
Next, calculate the moles of HCl:
Moles of HCl = mass / molar_mass
Moles of HCl = 656.25 g / (36.45 g/mol) = 18 mol
Now, calculate the molarity of the HCl solution:
Molarity = moles / volume
Molarity = 18 mol / 1.75 L = 10.3 M
For part (b), we have 22 g of HBr dissolved in 479 mL of solution. First, calculate the moles of HBr:
Moles of HBr = mass / molar_mass
Moles of HBr = 22 g / (80.91 g/mol) = 0.272 mol
Next, calculate the molarity of the HBr solution:
Molarity = moles / volume
Molarity = 0.272 mol / 0.479 L = 0.568 M
2Step 2: Calculate the concentration of H+ ions and pH
For strong acids like HCl and HBr, the concentration of H+ ions in the solution is equal to the molarity of the acid. Therefore, for part (a), [H+] = 10.3 M, and for part (b), [H+] = 0.568 M.
Now, calculate the pH of each solution:
pH = -log[H+]
For part (a):
pH = -log(10.3) ≈ -1.01
For part (b):
pH = -log(0.568) ≈ 0.25
3Step 3: Calculate the pH of the smaller volume in part (a)
We now need to find the pH of 0.175 L of the same HCl solution. Since the concentration of the solution remains the same, the [H+] is still equal to 10.3 M. Therefore, for the smaller volume of the solution:
pH = -log(10.3) ≈ -1.01
4Step 4: Summary
For part (a), [H+] = 10.3 M and pH ≈ -1.01 for both the 1.75 L solution and the 0.175 L solution.
For part (b), [H+] = 0.568 M and pH ≈ 0.25 for the solution with 22 g of HBr dissolved in 479 mL of water.
Key Concepts
Molarity of AcidConcentration of H+ IonspH Scale
Molarity of Acid
Molarity signifies the concentration of an acid or any solute in a solution, telling us the number of moles of the solute dissolved per liter of solution. To calculate it, you must divide the number of moles of the solute by the volume of the solution in liters. It is a crucial measure in chemistry because reactions depend on the precise molarity of reactants.
For instance, in our exercise, to find the molarity of hydrochloric acid (HCl), we first determined the mass of the solution using its density and volume, then worked out the mass and moles of HCl from its percentage composition. Dividing the moles of HCl by the volume of the solution in liters gave us the molarity. The exercise simplifies complex concepts by demonstration, making it an effective teaching tool.
For instance, in our exercise, to find the molarity of hydrochloric acid (HCl), we first determined the mass of the solution using its density and volume, then worked out the mass and moles of HCl from its percentage composition. Dividing the moles of HCl by the volume of the solution in liters gave us the molarity. The exercise simplifies complex concepts by demonstration, making it an effective teaching tool.
Concentration of H+ Ions
Understanding H+ Ion Concentration
When it comes to strong acids like HCl or hydrobromic acid (HBr), each molecule dissociates completely in water to release H+ ions (protons). The concentration of these ions is directly equivalent to the molarity of the acid in a given volume of water. This one-to-one correspondence simplifies calculations considerably and is crucial to understanding acid behavior.Our problem demonstrates this by directly equating the molarity of strong acids to the concentration of H+ ions. Knowing this concentration is fundamental as it directly influences the pH of the solution – a measure of acidity or basicity that has broad applications in chemistry, biology, environmental science, and medicine.
pH Scale
The pH scale is a logarithmic scale used to specify the acidity or basicity of an aqueous solution. It ranges from 0 to 14, with 7 being neutral, values below 7 acidic, and values above 7 basic. It's a direct way to understand the concentration of H+ ions in a solution without getting overwhelmed with large or small numbers – thanks to the logarithmic conversion.
To calculate pH, take the negative logarithm to the base 10 of the H+ ion concentration. Strong acids, as shown in our exercise, can have a pH less than zero, indicating very high acidity. Understanding the pH scale and its calculation is fundamental in various scientific fields, including chemistry, biology, and environmental science, because it affects every chemical process where water is a reactant or a product.
To calculate pH, take the negative logarithm to the base 10 of the H+ ion concentration. Strong acids, as shown in our exercise, can have a pH less than zero, indicating very high acidity. Understanding the pH scale and its calculation is fundamental in various scientific fields, including chemistry, biology, and environmental science, because it affects every chemical process where water is a reactant or a product.
Other exercises in this chapter
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