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Chapter 16: Problem 116

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 \(6.022 × 10^{13}\) H⁺(aq) ions in 1.0 mL of pure water at 25°C.

Step by step solution

01

Determine the ion product of water (Kw)

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

Find the concentration of H⁺ ions in pure water

In pure water, the concentrations of H⁺ ions and OH⁻ ions are equal, i.e., [H⁺] = [OH⁻]. We can use this fact along with the Kw value to solve for the concentration of H⁺ ions: \[[H^+] = \sqrt{K_w}\] \[[H^+] = \sqrt{1.0 × 10^{-14}}\] \[[H^+] = 1.0 × 10^{-7} \ \text{M}\] Now we have the concentration of H⁺ 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⁺ ions is given in moles per liter (M): \[1.0 \ \text{mL} = 1.0 × 10^{-3}\ \text{L}\]
04

Calculate the number of moles of H⁺ ions

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

Calculate the number of H⁺ ions

Finally, we can find the number of H⁺ ions using Avogadro's number (6.022 × 10²³ particles per mole): \[\text{Number of H}^+ \text{ ions} = n × \text{Avogadro's number}\] \[\text{Number of H}^+ \text{ ions} = (1.0 × 10^{-10} \ \text{moles}) × (6.022 × 10^{23} \ \text{ions/mol})\] \[\text{Number of H}^+ \text{ ions} = 6.022 × 10^{13} \ \text{ions}\] Thus, there are approximately \(6.022 × 10^{13}\) H⁺(aq) ions in 1.0 mL of pure water at 25°C.

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Most popular questions from this chapter

A particular sample of vinegar has a pH of \(2.90 .\) If acetic acid is the only acid that vinegar contains \(\left(K_{a}=1.8 \times 10^{-5}\right)\) calculate the concentration of acetic acid in the vinegar.

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