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)
calculate the number of H3O+ and OH- ions in 1.00 mL of pure water
2 answers
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.
(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.
1 answer