# 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

Mar 26, 2017

In pure water at $\text{298 K}$, the ${K}_{w}$ of water is ${10}^{- 14}$. Therefore, the equilibrium for the autoionization of water is:

${\text{H"_2"O"(l) rightleftharpoons "H"^(+)(aq) + "OH}}^{-} \left(a q\right)$

and its equilibrium constant expression is then:

${K}_{w} = {10}^{- 14} = \left[{\text{H"^(+)]["OH}}^{-}\right]$

But the ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ are 1:1, so $\left[{\text{H"^(+)] = ["OH}}^{-}\right]$. Therefore,

$\sqrt{{K}_{w}} = | \sqrt{{10}^{- 14}} | = | \sqrt{{\left({10}^{- 7}\right)}^{2}} | = {10}^{- 7} \text{M}$,

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"

$= 3 \times {10}^{- 8} \text{mols}$ ${\text{H}}^{+}$ or ${\text{OH}}^{-}$

Thus, the number of particles of each kind are:

$\textcolor{b l u e}{{N}_{{H}^{+}} = {N}_{O {H}^{-}}} = {n}_{{H}^{+}} \cdot {N}_{A}$

$= 3 \times {10}^{- 8} \text{mols" * 6.0221413 xx 10^(23) "particles/mol}$

$=$ $\textcolor{b l u e}{1.807 \times {10}^{16}}$ $\textcolor{b l u e}{\text{particles of}}$ $\textcolor{b l u e}{{\text{H}}^{+}}$ $\textcolor{b l u e}{\text{or}}$ $\textcolor{b l u e}{{\text{OH}}^{-}}$