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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. \(4.0 \mathrm{M} \mathrm{HNO}_{3}\)

Short Answer

Expert verified
In conclusion, the pH values of the given HNO3 solutions are: a. 1.70 b. -0.60 c. 11.21

Step by step solution

01

Since HNO3 is a strong acid and fully dissociates in water, the concentration of H+ ions is equal to the concentration of HNO3: \[2.0 \times 10^{-2} \mathrm{M}\]. #Step 2: Calculate the pH#

\(\mathrm{pH} = -\log_{10}[\mathrm{H^{+}}]\), where \([\mathrm{H^{+}}] = 2.0 \times 10^{-2} \mathrm{M}\). Therefore, the pH is approximately 1.70. #b. 4.0 M HNO3# #Step 1: Identify the concentration of H+ ions#
02

Since HNO3 is a strong acid and fully dissociates in water, the concentration of H+ ions is equal to the concentration of HNO3: \[4.0 \mathrm{M}\]. #Step 2: Calculate the pH#

\(\mathrm{pH} = -\log_{10}[\mathrm{H^{+}}]\), where \([\mathrm{H^{+}}] = 4.0 \mathrm{M}\). Therefore, the pH is approximately -0.60. (Note: While the pH is typically expected to be between 0 and 14, strong acid solutions prepared with molar concentrations greater than 1 can result in negative pH values. This indicates that the solution is extremely acidic.) #c. 6.2 x 10^-12 M HNO3# #Step 1: Identify the concentration of H+ ions#
03

Since HNO3 is a strong acid and fully dissociates in water, the concentration of H+ ions is equal to the concentration of HNO3: \[6.2 \times 10^{-12} \mathrm{M}\]. #Step 2: Calculate the pH#

\(\mathrm{pH} = -\log_{10}[\mathrm{H^{+}}]\), where \([\mathrm{H^{+}}] = 6.2 \times 10^{-12} \mathrm{M}\). Therefore, the pH is approximately 11.21. In conclusion, the pH values of the given HNO3 solutions are: a. 1.70 b. -0.60 c. 11.21

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Key Concepts

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

Strong Acids
Strong acids, like nitric acid (HNO₃), are characterized by their complete dissociation in water. When an acid dissociates completely, it means that every molecule of the acid breaks apart in water to form hydrogen ions (H⁺) and its corresponding anion. This results in a solution where the concentration of hydrogen ions is equal to the concentration of the acid itself.
Understanding strong acids is crucial because it simplifies the calculations for the concentration of hydrogen ions. When you know the concentration of a strong acid, you directly know the concentration of H⁺ ions in the solution.
Examples of strong acids include:
  • Hydrochloric acid (HCl)
  • Nitric acid (HNO₃)
  • Sulfuric acid (H₂SO₄)
These acids are used frequently in chemistry calculations due to their predictable behavior when dissolved.
Concentration of Hydrogen Ions
The concentration of hydrogen ions \([H⁺]\) in a solution is a direct indicator of the solution's acidity. In the case of strong acids, because they fully dissociate, the \([H⁺]\) is equal to the initial concentration of the acid.
For example, if the concentration of nitric acid in a solution is 2.0 x 10⁻² M, the concentration of hydrogen ions is also 2.0 x 10⁻² M. This simplification is quite beneficial for pH calculations.
Understanding this relationship helps in calculating the pH, which is the measure of acidity or alkalinity of a solution. Here's how it works:
  • Identify the \([H⁺]\)
  • Use the formula: \( ext{pH} = -\log_{10}[H⁺]\)
Note that this relationship holds true for strong acids due to their full dissociation, but may not apply to weak acids, which do not dissociate completely.
Negative pH Values
While the pH scale is typically thought to range between 0 and 14, in reality, there can be exceptions, especially with strong acids. When you have a very high concentration of \([H⁺]\) from a strong acid, the calculated pH can dip below zero, resulting in a negative pH.
Negative pH values indicate an extremely high acidity level and are possible with strong acids because of their complete dissociation in high concentrations.
For instance, if you have a solution with a molarity of 4.0 M HNO₃, the \([H⁺]\) is also 4.0 M, and the calculated pH is -0.60. This doesn't mean the concept of pH is broken; rather, it highlights the strength of acid in unique scenarios.
Remember:
  • Negative pH signifies very strong acids.
  • It reflects chemical environments outside typical laboratory conditions.
  • Approach these scenarios with caution, as they represent high reactivity.
Understanding this concept helps in better appreciation of the range of chemical reactions possible in different pH environments.

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Most popular questions from this chapter

A \(0.20 \mathrm{M}\) sodium chlorobenzoate \(\left(\mathrm{NaC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\) solution has a \(\mathrm{pH}\) of \(8.65 .\) Calculate the \(\mathrm{pH}\) of a \(0.20 \mathrm{M}\) chlorobenzoic acid \(\left(\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\) solution.

For the reaction of hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) in water. $$ \mathrm{H}_{2} \mathrm{NNH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{NNH}_{3}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ \(K_{\mathrm{b}}\) is \(3.0 \times 10^{-6}\). Calculate the concentrations of all species and the \(\mathrm{pH}\) of a \(2.0 \mathrm{M}\) solution of hydrazine in water.

Sodium azide \(\left(\mathrm{NaN}_{3}\right)\) is sometimes added to water to kill bacteria. Calculate the concentration of all species in a \(0.010 \mathrm{M}\) solution of \(\mathrm{NaN}_{3}\). The \(K_{\mathrm{a}}\) value for hydrazoic acid \(\left(\mathrm{HN}_{3}\right)\) is \(1.9 \times 10^{-5}\)

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?

Calculate the mass of \(\mathrm{HONH}_{2}\) required to dissolve in enough water to make \(250.0 \mathrm{~mL}\) of solution having a \(\mathrm{pH}\) of \(10.00\left(K_{\mathrm{b}}=\right.\) \(\left.1.1 \times 10^{-8}\right)\)

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