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Chapter 15: Problem 147

What is the \(\mathrm{pH}\) and the principal source of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions in \(1.0 \times 10^{-10} \mathrm{M} \mathrm{HCl} ?\) (Hint: The \(\mathrm{pH}\) of an acid solution can't exceed 7.) What is the \(\mathrm{pH}\) of \(1.0 \times 10^{-7} \mathrm{M} \mathrm{HCl}\) ?

Short Answer

Expert verified
For \(1.0 \times 10^{-10} \mathrm{M} \mathrm{HCl}\), \(\mathrm{pH} \approx 7\); principal source is water. For \(1.0 \times 10^{-7} \mathrm{M} \mathrm{HCl}\), \(\mathrm{pH} \approx 6.7\).

Step by step solution

01

Understand the problem

We need to find the \(\mathrm{pH}\) and the principal source of \(\mathrm{H}_{3}\mathrm{O}^{+} \) ions in two solutions: \(1.0 \times 10^{-10} \mathrm{M} \mathrm{HCl}\) and \(1.0 \times 10^{-7} \mathrm{M} \mathrm{HCl}\). The hint tells us that \(\mathrm{pH}\) cannot exceed 7 for an acidic solution.
02

Consider the autoionization of water

Water naturally ionizes into \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) ions: \(\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}^{+} + \mathrm{OH}^{-}\). The concentration of \(\mathrm{H}^{+}\) from water is \(1.0 \times 10^{-7}\ \mathrm{M}\). This must be considered when dealing with very dilute acid solutions.
03

Calculate the \(\mathrm{pH}\) of \(1.0 \times 10^{-10} \mathrm{M} \mathrm{HCl}\)

Since the concentration of \(\mathrm{HCl}\) is less than the \(\mathrm{H}^{+}\) from water, the ionic contribution from water is more significant. The total \(\mathrm{H}^{+}\) concentration is \(1.0 \times 10^{-7} + 1.0 \times 10^{-10}\ \mathrm{M} \approx 1.0 \times 10^{-7}\ \mathrm{M}\). Since the \(\mathrm{pH} = -\log([\mathrm{H}^{+}])\), \(\mathrm{pH} \approx 7\). The principal source of \(\mathrm{H}_{3}\mathrm{O}^{+} \) is water.
04

Calculate the \(\mathrm{pH}\) of \(1.0 \times 10^{-7} \mathrm{M} \mathrm{HCl}\)

In this case, \(\mathrm{H}^{+}\) from HCl and water are equal. Therefore, the total \(\mathrm{H}^{+}\) is \(2.0 \times 10^{-7}\ \mathrm{M}\). \(\mathrm{pH} = -\log(2.0 \times 10^{-7})\), which is slightly less than 7. Calculate precisely if needed for detailed understanding.

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Key Concepts

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

Autoionization of Water
Autoionization of water is a fascinating process where water molecules, even in pure form, naturally split into hydrogen ions (\( \mathrm{H}^{+} \)) and hydroxide ions (\( \mathrm{OH}^{-} \)).

This dynamic equilibrium can be represented by the equation:\[ \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}^{+} + \mathrm{OH}^{-} \]Even in the absence of added acids or bases, there are always \( 1.0 \times 10^{-7} \ \mathrm{M} \) of both \( \mathrm{H}^{+} \) and \( \mathrm{OH}^{-} \) ions at 25掳C.

This \( \mathrm{H}^{+} \) concentration in pure water establishes a neutral \( \mathrm{pH} \) level of 7.
  • It is crucial to consider this natural ionization in scenarios involving very dilute acid solutions.
  • This background ionization becomes a principal contributor to the overall hydrogen ion concentration when dealing with low concentrations of strong acids.
Dilute Acid Solutions
Dilute acid solutions contain a relatively low concentration of acid compared to their solvent.

For such solutions, the concentration of \( \mathrm{H}^{+} \) ions originating from water must not be neglected, especially if it exceeds or approaches the concentration from the acid itself.
  • This is primarily significant when the acid concentration falls below \( 1.0 \times 10^{-7} \ \mathrm{M} \).
  • In these cases, even a strong acid may not significantly impact the \( \mathrm{pH} \) compared to the autoionization of water.
This results in a \( \mathrm{pH} \) reading that reflects a balance of hydrogen ions from both the water and the acid.

When calculating \( \mathrm{pH} \) in such conditions, it is important to add these sources of \( \mathrm{H}^{+} \) together to get an accurate measurement.
HCl Acidity
Hydrochloric acid (\( \mathrm{HCl} \)) is a strong acid that dissociates fully in water to form hydrogen ions (\( \mathrm{H}^{+} \)) and chloride ions (\( \mathrm{Cl}^{-} \)).
  • When dissolved in water, each molecule of \( \mathrm{HCl} \) contributes directly to the increase in \( \mathrm{H}^{+} \) ion concentration.
  • Therefore, for higher concentrations, the \( \mathrm{pH} \) can be measured directly using the concentration of \( \mathrm{HCl} \).
However, when dilutions are excessive, the background \( \mathrm{H}^{+} \) from water's autoionization becomes more comparable in influence.

In such extremely dilute solutions as \( 1.0 \times 10^{-10} \ \mathrm{M}\) \( \mathrm{HCl} \), the dissociated ions contribute less significantly than those from water, skewing the \( \mathrm{pH} \) towards neutrality.

This characteristic distinguishes strong acids like \( \mathrm{HCl} \) from weaker ones, which do not dissociate completely.
Hydronium Ions
Hydronium ions (\( \mathrm{H}_{3}\mathrm{O}^{+} \)) form when a free hydrogen ion (\( \mathrm{H}^{+} \)) from an acid combines with a water molecule.

This is a common occurrence in aqueous solutions and gives rise to the acidic properties observed.
  • In simpler terms, whenever we measure or talk about \( \mathrm{H}^{+} \) concentration in solutions, they are actually present as the \( \mathrm{H}_{3}\mathrm{O}^{+} \) complex.
  • This transformation stabilizes the reactive hydrogen ion, making it easier to analyze and understand their behavior in water.
Consideration of \( \mathrm{H}_{3}\mathrm{O}^{+} \) ions is essential when determining the \( \mathrm{pH} \) of solutions, especially in those involving acids.

In any dilution or chemical process involving acids like \( \mathrm{HCl} \), it is these ions that directly influence the \( \mathrm{pH} \), tying together the contribution from both the acid and the solvent.

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Most popular questions from this chapter

During mining operations, the mineral pyrite \(\left(\mathrm{FeS}_{2}\right)\) is exposed to air and oxygen and reacts to produce sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\). The drainage from the Iron Mountain Mine in California was found to have \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=6.3 \mathrm{M}\). Calculate the \(\mathrm{pH}\) of the water seeping from the mine.

The pH of \(0.10 \mathrm{M}\) HOCl is \(4.23\). Calculate \(K_{\mathrm{a}}\) and \(\mathrm{pK}_{\mathrm{a}}\) for hypochlorous acid. and check your answers against the values given in Table \(15.2 .\)

Tartaric acid \(\left(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{6}\right)\) is a diprotic acid that plays an important role in lowering the \(\mathrm{pH}\) of wine to a level that bacteria cannot survive. Calculate the \(\mathrm{pH}\) of a \(0.50 \mathrm{M}\) tartaric acid solution \(\left(\mathrm{PK}_{a 1}=2.89 ; \mathrm{P} K_{\mathrm{a} 2}=4.40\right)\)

In the case of very weak acids, \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) from the dissociation of water is significant compared with \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) from the dissociation of the weak acid. The sugar substitute saccharin \(\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S}\right)\), for example, is a very weak acid having \(K_{u}=2.1 \times 10^{-12}\) and a solubility in water of \(348 \mathrm{mg} / 100 \mathrm{~mL}\). Calculate \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) in a saturated solution of saccharin. (Hint: Equilibrium equations for the dissociation of saccharin and water must be solved simultaneously.)

Give the formula for the conjugate acid of each of the following Bronsted- Lowry bases: (a) \(\mathrm{SO}_{3}^{2-}\) (b) \(\mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (d) \(\mathrm{OH}\) (e) \(\mathrm{HCO}_{3}^{-}\) (f) \(\mathrm{H}\)
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