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Calculate the number of \(\mathrm{H}^{+}(a q)\) ions in 1.0 \(\mathrm{mL}\) of pure water at \(25^{\circ} \mathrm{C} .\)

Short Answer

Expert verified
There are approximately \(6.022 脳 10^{13}\) H鈦(aq) ions in 1.0 mL of pure water at 25掳C.

Step by step solution

01

Determine the ion product of water (Kw)

At 25掳C, the ion product of water (Kw) is given as: \[K_w = [H^+][OH^-] = 1.0 脳 10^{-14}\] Where [H鈦篯 is the concentration of H鈦 ions and [OH鈦籡 is the concentration of OH鈦 ions.
02

Find the concentration of H鈦 ions in pure water

In pure water, the concentrations of H鈦 ions and OH鈦 ions are equal, i.e., [H鈦篯 = [OH鈦籡. We can use this fact along with the Kw value to solve for the concentration of H鈦 ions: \[[H^+] = \sqrt{K_w}\] \[[H^+] = \sqrt{1.0 脳 10^{-14}}\] \[[H^+] = 1.0 脳 10^{-7} \ \text{M}\] Now we have the concentration of H鈦 ions in the water.
03

Convert the volume of water from mL to Liters

We are given the volume of pure water as 1.0 mL. We need to convert this value to Liters, as the concentration of H鈦 ions is given in moles per liter (M): \[1.0 \ \text{mL} = 1.0 脳 10^{-3}\ \text{L}\]
04

Calculate the number of moles of H鈦 ions

Using the given concentration of H鈦 ions [H鈦篯 and the given volume of water, we can determine the number of moles of H鈦 ions (n): \[n = [H^+] 脳 \text{Volume of water in L}\] \[n = (1.0 脳 10^{-7} \ \text{M}) 脳 (1.0 脳 10^{-3} \text{L})\] \[n = 1.0 脳 10^{-10} \ \text{moles}\]
05

Calculate the number of H鈦 ions

Finally, we can find the number of H鈦 ions using Avogadro's number (6.022 脳 10虏鲁 particles per mole): \[\text{Number of H}^+ \text{ ions} = n 脳 \text{Avogadro's number}\] \[\text{Number of H}^+ \text{ ions} = (1.0 脳 10^{-10} \ \text{moles}) 脳 (6.022 脳 10^{23} \ \text{ions/mol})\] \[\text{Number of H}^+ \text{ ions} = 6.022 脳 10^{13} \ \text{ions}\] Thus, there are approximately \(6.022 脳 10^{13}\) H鈦(aq) ions in 1.0 mL of pure water at 25掳C.

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Key Concepts

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

Water Ionization
Water ionization is an important concept in chemistry, especially when dealing with the properties of pure water. When water undergoes ionization, it splits into hydrogen ions (H鈦) and hydroxide ions (OH鈦). This process can be represented by the following equilibrium equation:
2H鈧侽 鈬 H鈧僌鈦 + OH鈦
In pure water at 25掳C, this equilibrium creates equal concentrations of H鈦 and OH鈦, which are both 1.0 脳 10^{-7} M. This means that, even in its natural state, water contains a small number of ions.
Ionization of water is crucial because it contributes to the pH balance in aqueous solutions, making it a central focus in studies related to acids and bases. The fact that concentrations of both ions are equal in pure water simplifies calculations involved in determining pH and other acid-base phenomena.
- **Equilibrium Equation**: Demonstrates water's ionization into H鈦 and OH鈦. - **Concentration**: In pure water, both ions' concentrations are 1.0 脳 10^{-7} M. Understanding water ionization helps in comprehending broader chemical processes, like acid-base reactions and pH calculations.
Avogadro's Number
Avogadro's number is a key concept in chemistry, indicating the number of constituent particles, usually atoms or molecules, that are contained in one mole. The value of Avogadro's number is based on the number of atoms in 12 grams of carbon-12, which is 6.022 脳 10^{23} particles.
This number is vital for converting between the number of moles and the number of particles. For instance, when the number of moles of hydrogen ions is known, Avogadro's number allows you to calculate exactly how many ions are present.
- **Definition**: Avogadro's number is 6.022 脳 10^{23} particles per mole. - **Conversion Use**: It bridges the macroscopic scale of moles with the microscopic scale of individual ions or molecules. Avogadro's number acts as the link between measurable quantities of substances in chemistry and the vast number of tiny atoms or molecules they contain, greatly facilitating calculations involving chemical reactions and solutions.
Kw (Ion Product of Water)
The ion product of water, denoted as Kw, is a constant at a given temperature that signifies the product of the concentrations of hydrogen ions and hydroxide ions in water. At 25掳C, the value of Kw is 1.0 脳 10^{-14}.
Kw is essential because it captures the autoionization balance of water:
Kw = [H鈦篯[OH鈦籡
Since the concentration of H鈦 and OH鈦 in pure water is equal, you can deduce their concentrations by calculating the square root of Kw, which results in each being 1.0 脳 10^{-7} M under standard conditions.
- **Expression**: Kw = [H鈦篯[OH鈦籡 indicating the relationship between ion concentrations. - **Value**: At 25掳C, Kw is 1.0 脳 10^{-14}. Understanding Kw helps in various calculations in acid-base chemistry, as it allows chemists to determine respective ion concentrations in any aqueous solution, thereby assisting in pH computations and understanding the nature of solutions.

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Most popular questions from this chapter

An unknown salt is either \(\mathrm{KBr}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{KCN},\) or \(\mathrm{K}_{2} \mathrm{CO}_{3} .\) If a 0.100 \(\mathrm{M}\) solution of the salt is neutral, what is the identity of the salt?

Label each of the following as being a strong base, a weak base, or a species with negligible basicity. In each case write the formula of its conjugate acid, and indicate whether the conjugate acid is a strong acid, a weak acid, or a species with negligible acidity: \((\mathbf{a})\mathrm{CH}_{3} \mathrm{COO}^{-},(\mathbf{b}) \mathrm{HCO}_{3}^{-},(\mathbf{c}) \mathrm{O}^{2-},(\mathbf{d}) \mathrm{Cl}^{-},(\mathbf{e}) \mathrm{NH}_{3}\)

Deuterium oxide (\(\mathrm{D}_{2} \mathrm{O},\) where \(\mathrm{D}\) is deuterium, the hydrogen-2 isotope) has an ion-product constant, \(K_{w}\) , of \(8.9 \times 10^{-16}\) at \(20^{\circ} \mathrm{C}\). Calculate \(\left[\mathrm{D}^{+}\right]\) and \(\left[\mathrm{OD}^{-}\right]\) for pure (neutral) \(\mathrm{D}_{2} \mathrm{O}\) at this temperature.

Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water: (a) dimethylamine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH} ;(\mathbf{b})\) carbonate ion, \(\mathrm{CO}_{3}^{2-} ;(\mathbf{c})\) formate ion, \(\mathrm{CHO}_{2}^{-} .\)

Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water: (a) propylamine, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2} ;\) (b) monohydrogen phosphate ion, \(\mathrm{HPO}_{4}^{2-} ;(\mathbf{c})\) benzoate ion, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}.\)

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