Problem 39
Question
Calculate the \(\mathrm{pH}\) of an aqueous solution at \(25^{\circ} \mathrm{C}\) that is (a) \(1.02 \mathrm{M}\) in \(\mathrm{HI}\), (b) \(0.035 \mathrm{M}\) in \(\mathrm{HClO}_{4}\), and (c) \(1.5 \times 10^{-6} \mathrm{M}\) in \(\mathrm{HCl}\).
Step-by-Step Solution
Verified Answer
The pH values are approximately 0.0 for HI, 1.46 for HClO4, and 5.80 for HCl.
1Step 1: Understanding the Problem
The problem requires us to calculate the pH of three different aqueous solutions. Each solution contains a strong acid: HI, HClO4, and HCl. It's important to remember that strong acids fully dissociate in water.
2Step 2: Calculating pH for 1.02 M HI
Since HI is a strong acid, it completely dissociates in water. Thus, the concentration of H+ ions is equal to the concentration of HI.The pH is calculated using the formula: \[ \text{pH} = -\log[\text{H}^+] \]Substitute the concentration of H+ ions:\[ \text{pH} = -\log(1.02) = -0.009 \]Since pH is usually expressed with one decimal, pH ≈ 0.0.
3Step 3: Calculating pH for 0.035 M HClO4
HClO4 is also a strong acid and fully dissociates in water. This means that the concentration of H+ ions is equal to the concentration of HClO4.Calculate the pH:\[ \text{pH} = -\log[0.035] \approx 1.46 \]
4Step 4: Calculating pH for 1.5 x 10^-6 M HCl
Despite being a strong acid, the concentration of HCl (1.5 x 10^-6 M) is very low, which means we must also account for the autoinization of water which contributes \(1 \times 10^{-7}\) M of \(\text{H}^+\). The final \([\text{H}^+]\) is:\[ [\text{H}^+]_{\text{total}} = 1.5 \times 10^{-6} + 1 \times 10^{-7} = 1.6 \times 10^{-6} \]Calculate the pH:\[ \text{pH} = -\log[1.6 \times 10^{-6}] \approx 5.80 \]
5Step 5: Conclusion
For the given solutions, the calculated pH values are:
- The pH of the 1.02 M HI solution is approximately 0.0.
- The pH of the 0.035 M HClO4 solution is approximately 1.46.
- The pH of the 1.5 x 10^-6 M HCl solution is approximately 5.80.
Key Concepts
Strong AcidsAutoinization of WaterConcentration of H+ Ions
Strong Acids
When we talk about strong acids, it's essential to understand what makes an acid "strong." A strong acid is one that fully dissociates in water. This means that when you dissolve a strong acid in water, it breaks apart completely into its ions.
This complete dissociation allows for a straightforward pH calculation.
For the acids mentioned in the exercise:
Therefore, calculating the pH involves simply using this concentration.
This complete dissociation allows for a straightforward pH calculation.
For the acids mentioned in the exercise:
- HI (Hydroiodic acid)
- HClO₄ (Perchloric acid)
- HCl (Hydrochloric acid)
- H⁺ ions
- The corresponding negative ion
Therefore, calculating the pH involves simply using this concentration.
Autoinization of Water
Water is a fascinating substance due to its ability to autoionize. Autoinization of water refers to the process where water molecules dissociate into
When dealing with extremely diluted strong acids, like
For example, in the case of HCl with concentration being \(1.5 imes 10^{-6}\) M, the autoionization of water adds an additional \(1 imes 10^{-7}\) M to the total H⁺ concentration.
Ignoring this would slightly alter the accuracy of pH calculations.
- H⁺ ions
- OH⁻ ions
- \(1 imes 10^{-7}\) M
When dealing with extremely diluted strong acids, like
- a concentration close to or less than \(1 imes 10^{-7}\) M
For example, in the case of HCl with concentration being \(1.5 imes 10^{-6}\) M, the autoionization of water adds an additional \(1 imes 10^{-7}\) M to the total H⁺ concentration.
Ignoring this would slightly alter the accuracy of pH calculations.
Concentration of H+ Ions
One of the critical factors in pH calculations is understanding the concentration of H⁺ ions. The pH of a solution is directly related to the ul> [H⁺] through the formula:
\[ ext{pH} = -\log[ ext{H}^+] \]
When strong acids are present, calculating [H⁺] becomes much more direct. This is because the concentration of the acids themselves almost exactly equals the concentration of H⁺ ions.
However, with very low concentrations, such as in diluted strong acid solutions, we must also account for the additional H⁺ ions from the water's autoionization. Adding these concentrations provides a more complete picture of [H⁺], ensuring our pH calculations are precise.
The adjusted concentration is then used to find the pH, reflecting the total potential of hydrogen in the solution accurately.
\[ ext{pH} = -\log[ ext{H}^+] \]
When strong acids are present, calculating [H⁺] becomes much more direct. This is because the concentration of the acids themselves almost exactly equals the concentration of H⁺ ions.
However, with very low concentrations, such as in diluted strong acid solutions, we must also account for the additional H⁺ ions from the water's autoionization. Adding these concentrations provides a more complete picture of [H⁺], ensuring our pH calculations are precise.
The adjusted concentration is then used to find the pH, reflecting the total potential of hydrogen in the solution accurately.
Other exercises in this chapter
Problem 37
Why are ionizations of strong acids and strong bases generally not treated as equilibria?
View solution Problem 38
Calculate the \(\mathrm{pH}\) of an aqueous solution at \(25^{\circ} \mathrm{C}\) that is (a) \(0.12 \mathrm{M}\) in \(\mathrm{HCl}\), (b) \(2.4 \mathrm{M}\) in
View solution Problem 40
Calculate the concentration of \(\mathrm{HBr}\) in a solution at \(25^{\circ} \mathrm{C}\) that has a pH of (a) 0.12 , (b) \(2.46,\) and \((\mathrm{c}) 6.27\).
View solution Problem 41
Calculate the concentration of \(\mathrm{HNO}_{3}\) in a solution at \(25^{\circ} \mathrm{C}\) that has a \(\mathrm{pH}\) of \((\mathrm{a}) 4.21,(\mathrm{~b}) 3
View solution