Question

How to calculate the number of H+ and OH- ions in this question?

calculate the number of H+ and OH- ions in 300ml of pure water. on a temperature of 298k

Answer 1

In pure water at `"298 K"`, the `K_w` of water is `10^(-14)`. Therefore, the equilibrium for the autoionization of water is:

`"H"_2"O"(l) rightleftharpoons "H"^(+)(aq) + "OH"^(-)(aq)`

and its equilibrium constant expression is then:

`K_w = 10^(-14) = ["H"^(+)]["OH"^(-)]`

But the `"H"^(+)` and `"OH"^(-)` are 1:1, so `["H"^(+)] = ["OH"^(-)]`. Therefore,

`sqrt(K_w) = |sqrt(10^(-14))| = |sqrt((10^(-7))^2)| = 10^(-7) "M"`,

and we have the concentrations

`["H"^(+)] = ["OH"^(-)] = 10^(-7) "M"`.

So, we can calculate the mols of either one.

`n_(H^(+)) = ["OH"^(-)] xx "300 mL" xx "1 L"/"1000 mL"`

`= 3 xx 10^(-8) "mols"` `"H"^(+)` or `"OH"^(-)`

Thus, the number of particles of each kind are:

`color(blue)(N_(H^(+)) = N_(OH^(-))) = n_(H^(+))*N_A`

`= 3 xx 10^(-8) "mols" * 6.0221413 xx 10^(23) "particles/mol"`

`=` `color(blue)(1.807 xx 10^(16))` `color(blue)("particles of")` `color(blue)("H"^(+))` `color(blue)("or")` `color(blue)("OH"^(-))`