Question

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

Short answer

There are approximately \(6.022 \times 10^{13}\) H+ ions in 1.0 mL of pure water at 25°C.

Step-by-step solution

1. Find the ion product constant (Kw) at 25°C

At 25°C, the ion product constant for water is given as: \(K_w = [H^+][OH^-] = 1.0 \times 10^{-14}\)

2. Calculate the concentration of H+ ions in pure water

In pure water, the concentrations of H+ and OH- ions are equal, as the number of H+ ions formed is equal to the number of OH- ions. Therefore: \([H^+] = [OH^-]\) We can then rewrite Kw as: \[K_w = [H^+]^2\] Now, solve for [H+]: \[[H^+] = \sqrt{K_w} = \sqrt{1.0 \times 10^{-14}}\]

3. Determine the concentration of H+ ions

Calculating the square root: \[[H^+] = 1.0 \times 10^{-7} \,M\]

4. Calculate the number of H+ ions in 1.0 mL of pure water

To do this, we will first convert the volume from mL to L: \[1.0\, mL = 1.0 \times 10^{-3} L\] Now, we can use the concentration formula (M = mol/L) to determine the number of moles of H+ ions in the given volume: \[moles\,of\,H^+ = [H^+] \times volume\] \[moles\,of\,H^+ = (1.0 \times 10^{-7}\,M) \times (1.0 \times 10^{-3}\,L)\] Calculating the product: \[moles\,of\,H^+ = 1.0 \times 10^{-10}\,mol\]

5. Determine the number of H+ ions

Finally, multiply the number of moles with Avogadro's number to find the total number of H+ ions: \[number\,of\,H^+ = moles\,of\,H^+ \times Avogadro's\,number\] \[number\,of\,H^+ = (1.0 \times 10^{-10}\,mol) \times (6.022 \times 10^{23}\,ion/mol)\] Calculating the product: \[number\,of\,H^+ \approx 6.022 \times 10^{13}\,ion\] There are approximately \(6.022 \times 10^{13}\) H+ ions in 1.0 mL of pure water at 25°C.