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.


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  • $\begingroup$ @Karl: I meant number of protons in the molecule natriumhydroxide. I changed it. $\endgroup$ – Anonymous196 Sep 30 '18 at 18:18

You need more than a hint. You need a hard shove...

The overall notion here is that water dissociates

$$\ce{H2O <=> H+ + OH-}$$

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.

$$\ce{NaOH(solid) <=> Na+(aq) + OH-(aq)}$$

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

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

  • $\begingroup$ Ok, I didn't see that coming. I didn't yet cover that part of chemistry. I'm familiar with the formulas but that's just because I read it from Wikipedia. I'll cover that part now. Thx for the answer though. I think it would be enough if you'd say use pH and the solubility product. My book I used a while ago has a chapter about it. $\endgroup$ – Anonymous196 Sep 30 '18 at 19:51
  • $\begingroup$ Hmmm... water doesn't have a "solubility product." There is no precipitate. $K_w$ is known as the (a) ionization constant, (b) dissociation constant, (c) self-ionization constant, or (d) ionic product of water, but never as the solubility product. $\endgroup$ – MaxW Sep 30 '18 at 20:05
  • $\begingroup$ I mean my book discusses that term in that chapter of acids and bases. $\endgroup$ – Anonymous196 Sep 30 '18 at 20:17

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