calculate the number of H3O+ and OH- ions in 1.00 mL of pure water
It is about 10^-7 of the number of molecules of H2O present. 1.0 ml has a mass of 1.0 g, which is 1/18 of a mole, for water.
Therefore the answer is
10^-7*(1/18)*(Avogadro's number)
I have struggled with this problem trying to get two methods to work out to the same answer and I can't do it (easily). As I see it, the problem should be done this way.
(H3O^+) = (OH^-) = 1 x 10^-7 moles/L in pure water.
1 mL is 0.001 L; therefore,
1 x 10^-7 moles/L x 0.001 L x 6.02 x 10^23 ions/mole = # H3O^+ = #OH^- in 1 mL.
If we examine the units for the other way of doing it,
1/18 = moles and
1 g x (1 mole/18 g) x (6.02 x 10^23 molecules/mole) = # molecules water. So far so good.
But if we multiply # molecules water x (1 x 10^-7 moles/L) we don't get # H3O^+/1 mL. I believe the problem is that 1 x 10^-7 is number moles H3O^+ in 55.56 moles H2O and if that factor is included, the answers are the same. Check my thinking.
To calculate the number of H3O+ and OH- ions in 1.00 mL of pure water, we need to first determine the concentration of H3O+ and OH- ions in water.
In pure water at 25°C, the concentration of H3O+ ions, also known as the hydronium ions, is equal to the concentration of OH- ions. This concentration is often represented by the symbol "Kw" and has a constant value of 1.00 x 10^-14 mol/L.
Since we are dealing with a volume of 1.00 mL, we need to convert this volume to liters by dividing it by 1000 (1 mL = 0.001 L).
Volume of pure water = 1.00 mL = 0.001 L
Now, we can calculate the concentration of H3O+ and OH- ions using the constant value of Kw.
[H3O+] = [OH-] = Kw^(1/2)
[H3O+] = [OH-] = (1.00 x 10^-14 mol/L)^(1/2)
[H3O+] = [OH-] = 1.00 x 10^-7 mol/L
Now that we have the concentration of both H3O+ and OH- ions in mol/L, we can calculate the number of ions using the given volume.
Number of H3O+ ions = [H3O+] x Volume
Number of OH- ions = [OH-] x Volume
Number of H3O+ ions = 1.00 x 10^-7 mol/L x 0.001 L
Number of OH- ions = 1.00 x 10^-7 mol/L x 0.001 L
Number of H3O+ ions = 1.00 x 10^-10 mol
Number of OH- ions = 1.00 x 10^-10 mol
Therefore, in 1.00 mL of pure water, there are 1.00 x 10^-10 mol (or the same number) of H3O+ ions and OH- ions.
To calculate the number of H3O+ and OH- ions in pure water, you need to know the concentration of these ions. In pure water, at 25 degrees Celsius, the concentration of H3O+ and OH- ions is 1.0 x 10^-7 M.
To find the number of moles of H3O+ or OH- ions, you can multiply the concentration by the volume of the solution in liters. Since you have 1.00 mL of water, you need to convert it to liters by dividing by 1000:
1.00 mL ÷ 1000 = 0.001 L
Next, you can calculate the number of moles of H3O+ or OH- ions using the formula:
moles = concentration × volume
For H3O+ ions:
moles of H3O+ = (1.0 x 10^-7 M) × (0.001 L)
moles of H3O+ = 1.0 x 10^-10 moles
Similarly, for OH- ions:
moles of OH- = (1.0 x 10^-7 M) × (0.001 L)
moles of OH- = 1.0 x 10^-10 moles
Therefore, in 1.00 mL of pure water, there are approximately 1.0 x 10^-10 moles of both H3O+ and OH- ions.