Problem 89
Question
Which has the larger concentration of hydrogen ions, \(0.015 \mathrm{M} \mathrm{HCl}\) or a hydrochloric acid solution with a \(\mathrm{pH}\) of \(1.2 ?\)
Step-by-Step Solution
Verified Answer
A solution with a pH of 1.2 has a higher hydrogen ion concentration (0.0631 M) than a 0.015 M HCl solution.
1Step 1: Understand Hydrogen Ion Concentration
Hydrogen ion concentration in a solution is given by the molarity of hydrogen ions. For hydrochloric acid (HCl), a strong acid, it dissociates completely, meaning the hydrogen ion concentration is equal to the molarity of the HCl solution. Thus, for a 0.015 M HCl solution, the concentration of hydrogen ions is 0.015 M.
2Step 2: Calculate Hydrogen Ion Concentration from pH
To find the concentration of hydrogen ions from the pH, use the formula \([H^+] = 10^{- ext{pH}}\). If the pH is 1.2, then \([H^+] = 10^{-1.2}\). Calculate this to get \([H^+] \approx 0.0631 \text{ M}\).
3Step 3: Compare the Hydrogen Ion Concentrations
Now compare the two concentrations calculated. The 0.015 M HCl solution has a hydrogen ion concentration of 0.015 M, and the solution with pH 1.2 has a concentration of approximately 0.0631 M.
Key Concepts
pH calculationmolaritystrong acidsdissociation of hydrochloric acid
pH calculation
Calculating the pH of a solution is crucial for understanding its acidity or basicity. The pH scale ranges from 0 to 14, with 7 being neutral. Values less than 7 indicate acidity, and those greater than 7 indicate basicity. The formula to calculate pH is \[ \text{pH} = -\log[H^+] \] where \([H^+]\) is the concentration of hydrogen ions in moles per liter (M). This logarithmic scale helps in managing the wide range of hydrogen ion concentrations possible in solutions.
For instance, a solution with a hydrogen ion concentration of 0.01 M has a pH of 2, while a highly acidic solution with a \([H^+]\) of 0.1 M has a pH of 1. Calculating pH helps in predicting the chemical behavior of a substance and is a key part of analytical chemistry.
Always ensure to use a calculator when dealing with logarithms to obtain precise results.
For instance, a solution with a hydrogen ion concentration of 0.01 M has a pH of 2, while a highly acidic solution with a \([H^+]\) of 0.1 M has a pH of 1. Calculating pH helps in predicting the chemical behavior of a substance and is a key part of analytical chemistry.
Always ensure to use a calculator when dealing with logarithms to obtain precise results.
molarity
Molarity is a way to express the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution, symbolized as M. \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}\] This unit is integral to many chemical calculations as it helps determine how substances interact in a solution.
For example, when dealing with hydrochloric acid solutions, if you have 0.015 M HCl, this means there are 0.015 moles of HCl in every liter of the solution.
Understanding molarity is essential for calculating how reactions occur in solutions and predicting the extent of reactions.
For example, when dealing with hydrochloric acid solutions, if you have 0.015 M HCl, this means there are 0.015 moles of HCl in every liter of the solution.
Understanding molarity is essential for calculating how reactions occur in solutions and predicting the extent of reactions.
strong acids
Strong acids are a category of acids that dissociate completely in solution. This complete dissociation means that they release all their available hydrogen ions into the solution, increasing the solution’s acidity.
Some common strong acids include:
This characteristic of strong acids makes them useful in various industrial applications where complete ionization and a predictable pH are required.
Some common strong acids include:
- Hydrochloric acid (HCl)
- Sulfuric acid (H₂SO₄)
- Nitric acid (HNO₃)
This characteristic of strong acids makes them useful in various industrial applications where complete ionization and a predictable pH are required.
dissociation of hydrochloric acid
The dissociation of hydrochloric acid (HCl) in water is a straightforward process. Since HCl is a strong acid, it dissociates completely in aqueous solutions.
This dissociation can be represented by the equation: \[ \text{HCl (aq)} \rightarrow \text{H}^+ \text{(aq)} + \text{Cl}^- \text{(aq)} \] In this reaction, each molecule of HCl releases one hydrogen ion \(\text{H}^+\) into the solution, resulting in equal concentrations of \(\text{H}^+\) and \(\text{Cl}^-\) ions.
This complete dissociation makes it easy to calculate the concentration of hydrogen ions in an HCl solution if you know its molarity. For instance, a 0.015 M solution of HCl will have a \([H^+]\) of 0.015 M. Understanding this process is key to mastering concepts such as pH and other related chemical equilibria in solutions.
This dissociation can be represented by the equation: \[ \text{HCl (aq)} \rightarrow \text{H}^+ \text{(aq)} + \text{Cl}^- \text{(aq)} \] In this reaction, each molecule of HCl releases one hydrogen ion \(\text{H}^+\) into the solution, resulting in equal concentrations of \(\text{H}^+\) and \(\text{Cl}^-\) ions.
This complete dissociation makes it easy to calculate the concentration of hydrogen ions in an HCl solution if you know its molarity. For instance, a 0.015 M solution of HCl will have a \([H^+]\) of 0.015 M. Understanding this process is key to mastering concepts such as pH and other related chemical equilibria in solutions.
Other exercises in this chapter
Problem 87
You have a bottle of solid \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and a \(500.0-\mathrm{mL}\). volumetric flask. Explain how you would make a \(0.20 \mathrm{M}\) s
View solution Problem 88
You have 0.500 mol of \(\mathrm{KCl}\), some distilled water, and a \(250 .-\mathrm{mL}\) volumetric flask. Describe how you would make a \(0.500 \mathrm{M}\) s
View solution Problem 90
What volume of \(0.054 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is required to react completely with \(1.56 \mathrm{g}\) of \(\mathrm{KOH} ?\)
View solution Problem 92
Ammonium sulfide, (NH,) \(_{2} \mathrm{S}\), reacts with \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) to produce HgS and \(\mathrm{NH}_{4} \mathrm{NO}_{3}\)
View solution