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

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

Step-by-step solution

Determine the ion product constant (Kw) for water at 25°C

At 25°C, the ion product constant for water (Kw) is \(1.0 \times 10^{-14}\) . In pure water, it is equal to the concentration of H+ ions multiplied by the concentration of hydroxide (OH-) ions, i.e., \[K_w = [ H ^ + ][ OH ^ - ]\]

Find the concentration of H+ ions

In pure water, the concentration of H+ ions is equal to the concentration of hydroxide (OH-) ions. Since we know the ion product constant, we can solve for the concentration of H+ ions: \[[ H ^ + ] [ OH ^ - ] = 1.0 \times 10 ^{-14}\] \[[ H ^ + ] ^ 2 = 1.0 \times 10 ^{-14}\] \[[ H ^ + ] = \sqrt{1.0 \times 10 ^{-14}} = 1.0 \times 10^{-7} \mathrm{M}\]

Convert mL to L

We need to convert the volume of water from mL to L to find the number of moles of H+ ions in the water: \[1.0 \mathrm{mL} = 1.0 \times 10 ^{-3} \mathrm{L}\]

Calculate the number of moles of H+ ions

Now that we have the concentration of H+ ions and the volume in liters, we can find the number of moles of H+ ions in the water: \[[ H ^ + ] = \frac{\text{number of moles of H}^{+}}{\text{volume in L}}\] \[\text{number of moles of H}^{+} = [ H ^ + ] \times \text{volume}\] \[\text{number of moles of H}^{+} = (1.0 \times 10^{-7} \mathrm{M}) (1.0 \times 10^{-3} \mathrm{L}) = 1.0 \times 10^{-10}\mathrm{mol}\]

Calculate the number of H+ ions using Avogadro's number

Now we can use Avogadro's number, the number of particles in a mole, to find the number of H+ ions: \[\text{number of H}^{+} \text{ions} = \text{number of moles of H}^{+} \times \text{Avogadro's number}\] \[\text{number of H}^{+} \text{ions} = (1.0 \times 10^{-10}\text{mol})(6.022 \times 10^{23} \text{ions/mol}) =6.022 \times 10^{13} \text{ions}\] In conclusion, there are \(6.022 \times 10^{13}\) H+ ions in 1.0 mL of pure water at 25°C.