I have an experiment where I drop 1 gram of lab-concentration Sodium Hydroxide into a litre sea water (pH 8.1). What should be the expected change in pH? Would this change be constant? i.e if I was to drop 2 grams of NaOH into the sea water, would the change double?

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    $\begingroup$ As I understand, you're looking for a more or less precise calculation? If so, we need to know the volume of said "little sea water" since the pH would depend on the resulting concentration of the sodium hydroxide. Also, what do you mean by "lab-concentration" sodium hydroxide?? $\endgroup$ – ChemistryHelpCenter Jul 22 '16 at 0:46
  • $\begingroup$ @ChemistryHelpCenter thanks a lot for replying. The volume is a litre of sea water. Lab concentration means about pH 13-13.4 $\endgroup$ – SuperNinja741 Does Gaming Jul 22 '16 at 0:53

If we know that the pH is 13 (let's take it as 13 for the simplicity), then we can back-calculate the concentration of sodium hydroxide in this solution from $\ce{14=pH + pOH}$. This gives us pOH=1, thus the concentration of sodium hydroxide of 0.1M. Since 0.1M sodium hydroxide is hardly a very concentrated solution, we can safely assume that 1 gram of the solution is going to be roughly 1 mL. Thus, with each 1 gram of the lab solution you're adding about 0.0001 moles of $\ce{NaOH}$ into your 1 liter of sea water giving you $1*10^{-4}$M solution with the pH = 10.

Using the same trick to calculate the concentration of $\ce{OH-}$ in your sea water as we did for the stock sodium hydroxide solution, we get about $1.25*10^{-6}$M, which is negligibly small compared to $1*10^{-4}$M solution that you get after adding sodium hydroxide to it. Thus, after 1 gram of sodium hydroxide in a liter of sea water your pH should go from 8.1 to about 10. If you add 2 grams, you'll get a $2*10^{-4}$M solution, which yields pH of about 10.3.


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