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Chapter 14: Problem 55

Calculate the \(\mathrm{pH}\) of each of the following solutions of a strong acid in water. a. \(0.10 \mathrm{M} \mathrm{HCl}\) c. \(1.0 \times 10^{-11} \mathrm{M} \mathrm{HCl}\) b. \(5.0 \mathrm{M} \mathrm{HCl}\)

Short Answer

Expert verified
The pH values for the given strong acid solutions are as follows: a. \(0.10 \mathrm{M} \mathrm{HCl}\): pH = 1.00 b. \(1.0 \times 10^{-11} \mathrm{M} \mathrm{HCl}\): pH = 11 c. \(5.0 \mathrm{M} \mathrm{HCl}\): pH ≈ -0.30

Step by step solution

01

Identify the concentration of hydrogen ions

Since HCl is a strong acid, it completely dissociates in water and releases an equal concentration of hydrogen ions (H+). In this case, the concentration of H+ ions is equal to the concentration of HCl: \[[\mathrm{H}^+] = 0.10 \mathrm{M}\]
02

Calculate the pH

Now we can use the pH formula to calculate the pH value: \[pH = -\log_{10}[\mathrm{H}^+] = -\log_{10}(0.10) = 1.00\] The pH of a \(0.10 \mathrm{M} \mathrm{HCl}\) solution is 1.00. b. Calculate the pH of a \(1.0 \times 10^{-11} \mathrm{M} \mathrm{HCl}\) solution
03

Identify the concentration of hydrogen ions

Since HCl is a strong acid, it completely dissociates in water, releasing an equal concentration of H+ ions. In this case, \[[\mathrm{H}^+] = 1.0 \times 10^{-11} \mathrm{M}\]
04

Calculate the pH

Now we can use the pH formula to calculate the pH value: \[pH = -\log_{10}[\mathrm{H}^+] = -\log_{10}(1.0 \times 10^{-11}) = 11\] The pH of a \(1.0 \times 10^{-11} \mathrm{M} \mathrm{HCl}\) solution is 11. c. Calculate the pH of a \(5.0 \mathrm{M} \mathrm{HCl}\) solution
05

Identify the concentration of hydrogen ions

Since HCl is a strong acid, it completely dissociates in water, releasing an equal concentration of H+ ions. In this case, \[[\mathrm{H}^+] = 5.0 \mathrm{M}\]
06

Calculate the pH

Now we can use the pH formula to calculate the pH value: \[pH = -\log_{10}[\mathrm{H}^+] = -\log_{10}(5.0) \approx -0.30\] The pH of a \(5.0 \mathrm{M} \mathrm{HCl}\) solution is approximately -0.30.

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Key Concepts

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

Strong Acid Dissociation
Understanding strong acid dissociation is crucial for grasping the basics of pH calculation. A strong acid, such as hydrochloric acid (HCl), dissociates completely in an aqueous solution. This means that almost every molecule of the acid splits into its constituent ions when dissolved in water. For HCl, it dissociates into hydrogen ions ( [H+]) and chloride ions ( [Cl-]).

Complete dissociation is an important concept because it indicates that the concentration of hydrogen ions is equal to the concentration of the acid. For example, if you have a 0.10 M solution of HCl, you can assume that the concentration of hydrogen ions is also 0.10 M.

This characteristic of strong acids allows us to easily calculate the pH of their solutions using the pH formula \[pH = -\log_{10}[H^+]\]. Knowing the concentration of hydrogen ions directly gives you a simple way to find the pH.
Hydrochloric Acid
Hydrochloric acid is one of the most common strong acids encountered in many chemistry problems and applications. It is formed when hydrogen chloride (HCl) gas is dissolved in water. Because it is a strong acid, it ionizes completely into hydrogen ions ( [H+] ) and chloride ions upon dissolution.

HCl is widely used in laboratories and industries due to its reactivity and ability to effectively donate protons to other substances. However, it is also found naturally, for example, in the gastric juices of the stomach, where it aids in digestion.

The complete dissociation of HCl in water results in a straightforward relationship between the molarity of the acid and the concentration of [H+] . This simplifies pH calculations, as knowing the initial concentration of HCl allows for a direct calculation of the hydrogen ion concentration.
Hydrogen Ion Concentration
Hydrogen ion concentration is central to understanding both acidity and pH levels in chemical solutions. The concentration of hydrogen ions ( [H+]) determines the acidity of a solution: the higher the concentration, the more acidic the solution.

pH is a logarithmic scale used to measure how acidic or basic a solution is. It is inversely related to the hydrogen ion concentration, given by the formula \[pH = -\log_{10}[H^+]\]. This implies that even small changes in hydrogen ion concentration can result in significant changes in pH values.

When dealing with strong acids like HCl, calculating hydrogen ion concentration is straightforward because the concentration of [H+] is directly equal to the concentration of the acid due to complete dissociation. Understanding this direct link simplifies calculations and helps predict the pH accurately for solutions of strong acids.

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Most popular questions from this chapter

Hemoglobin (abbreviated \(\mathrm{Hb}\) ) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemoglobin molecule contains four iron atoms that are the binding sites for \(\mathrm{O}_{2}\) molecules. The oxygen binding is pH-dependent. The relevant equilibrium reaction is $$\mathrm{HbH}_{4}^{4+}(a q)+4 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)+4 \mathrm{H}^{+}(a q)$$ Use Le Châtelier's principle to answer the following. a. What form of hemoglobin, \(\mathrm{HbH}_{4}{ }^{4+}\) or \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}\), is favored in the lungs? What form is favored in the cells? b. When a person hyperventilates, the concentration of \(\mathrm{CO}_{2}\) in the blood is decreased. How does this affect the oxygenbinding equilibrium? How does breathing into a paper bag help to counteract this effect? (See Exercise 148.) c. When a person has suffered a cardiac arrest, injection of a sodium bicarbonate solution is given. Why is this necessary?

A sample containing \(0.0500 \mathrm{~mol} \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is dissolved in enough water to make \(1.00 \mathrm{~L}\) of solution. This solution contains hydrated \(\mathrm{SO}_{4}{ }^{2-}\) and \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\) ions. The latter behaves as an acid: $$\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q) \rightleftharpoons \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}^{2+}(a q)+\mathrm{H}^{+}(a q)$$ a. Calculate the expected osmotic pressure of this solution at \(25^{\circ} \mathrm{C}\) if the above dissociation is negligible. b. The actual osmotic pressure of the solution is \(6.73\) atm at \(25^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{a}}\) for the dissociation reaction of \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\). (To do this calculation, you must assume that none of the ions goes through the semipermeable membrane. Actually, this is not a great assumption for the tiny \(\mathrm{H}^{+}\) ion.)

Calculate the \(\mathrm{pH}\) of a \(0.10 \mathrm{M} \mathrm{CoCl}_{3}\) solution. The \(K_{\mathrm{a}}\) value for \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is \(1.0 \times 10^{-5}\).

Calculate the \(\mathrm{pH}\) of a \(0.200 \mathrm{M}\) solution of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\). Hint: \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\) is a salt composed of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{F}^{-}\) ions. The prin- cipal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons\) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}(a q)+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\)

Calculate the \(\mathrm{pH}\) of each of the following solutions containing a strong acid in water. a. \(2.0 \times 10^{-2} \mathrm{M} \mathrm{HNO}_{3}\) c. \(6.2 \times 10^{-12} \mathrm{M} \mathrm{HNO}_{3}\) b. \(4.0 \mathrm{M} \mathrm{HNO}_{3}\)
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