How to determine the proton concentration in the following solution in mol/L?
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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

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