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Chapter 16: Problem 39

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 \(\mathrm{HCl}\).

Short Answer

Expert verified
The pH values are approximately 0.0 for HI, 1.46 for HClO4, and 5.80 for HCl.

Step by step solution

01

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.
02

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 鈮� 0.0.
03

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:\[ \text{pH} = -\log[0.035] \approx 1.46 \]
04

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:\[ \text{pH} = -\log[1.6 \times 10^{-6}] \approx 5.80 \]
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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:
  • HI (Hydroiodic acid)
  • HClO鈧� (Perchloric acid)
  • HCl (Hydrochloric acid)
They all behave similarly under typical conditions because of their strong nature. Breaking fully into
  • H鈦� ions
  • The corresponding negative ion
results in a direct relationship between the acid's initial concentration and the H鈦� ions in the solution.
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
  • H鈦� ions
  • OH鈦� ions
even without any added substances. At 25掳C, this process provides a constant H鈦� concentration of approximately
  • \(1 imes 10^{-7}\) M
under neutral conditions.
When dealing with extremely diluted strong acids, like
  • a concentration close to or less than \(1 imes 10^{-7}\) M
this autoionization must be considered.
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鈦� 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鈦� ions. The pH of a solution is directly related to the ul>
  • 摆贬鈦篯
    • through the formula:
    \[ ext{pH} = -\log[ ext{H}^+] \]
    When strong acids are present, calculating 摆贬鈦篯 becomes much more direct. This is because the concentration of the acids themselves almost exactly equals the concentration of H鈦� ions.
    However, with very low concentrations, such as in diluted strong acid solutions, we must also account for the additional H鈦� ions from the water's autoionization. Adding these concentrations provides a more complete picture of 摆贬鈦篯, 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.

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    Most popular questions from this chapter

    In terms of orbitals and electron arrangements, what must be present for a molecule or an ion to act as a Lewis acid (use \(\mathrm{H}^{+}\) and \(\mathrm{BF}_{3}\) as examples)? What must be present for a molecule or ion to act as a Lewis base (use \(\mathrm{OH}^{-}\) and \(\mathrm{NH}_{3}\) as examples)?

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