How to determine the proton concentration in the following solution in mol/L? A $\pu{0.8L}$ solution contains $\pu{0.60 g }\ce{ NaOH}$ in water. Then the above question follows. 


It's been a long time that I made some chemistry exercises. Now I'm stuck with a couple of hesitations. 

I know one needs to calculate the number of protons of the molecule sodium hydroxide. But do you need to take water into account or not? I'm thinking not but then the question asks for the solution. So I probably should do that which gets me to $\pu{29 g mol-1}$. Otherwise it is $\pu{20 g mol-1}$. 

So that I would divide by $\pu{0.60g}$ which gives me for the first one $\pu{48 mol}$ and the second one $\pu{33 mol}$. Then I'd divide it by $\pu{8.0L}$. 

I probably should have a larger answer than what I've done so it is not correct but I don't know exactly how I should work this out. 

I'd appreciate a hint.
You need more than a hint. You need a hard shove...

The overall notion here is that water dissociates 

$$\ce{H2O &lt;=&gt; H+ + OH-}$$

and that the equilibrium is given by

$$K_w = 1.00\times10^{-14} = \ce{[H+][OH-]}$$

where $\ce{[H+]}$ is the molar concentration of $\ce{H+}$ ions and $\ce{[OH-]}$ is the molar concentration of $\ce{OH-}$ ions.

NOTE: The pH of a solution is defined as the negative log of the $\ce{H+}$ ion concentration, and pOH of a solution is defined as the negative log of the $\ce{OH-}$ ion concentration. So

$$14 = \text{pH + pOH}$$

Sodium hydroxide is a strong base, so the assumption is that it completely dissociates. 

$$\ce{NaOH(solid) &lt;=&gt; Na+(aq) + OH-(aq)}$$

The molecular mass of sodium hydroxide is 40.00 grams/mole.

$$\ce{[OH-]} = \dfrac{0.60\text{ g}}{40.00\text{ g/mole}\times0.80\text{ liter}} = 0.01875\text{ mole/liter}$$

Rearranging the $K_w$ equation

$$\ce{[H+]} = \dfrac{K_w}{\ce{[OH-]}} = \dfrac{1.00\times10^{-14}}{1.875\times10^{-2}} = 5.3\times10^{-13}\text{ mole/liter}$$