A particular beaker contains 2.38x10-3 M NaOH. The hydrogen ion concentration , [H+], in this beaker is?



Answer:
This is a strong base so it completely dissociates into Na+ + OH-.

The pOH is going to be -log (2.38x10^-3) = 2.62

The pH is going to be 14-2.62 = 11.37

10^-11.37 = [H+] = 4.07x10^-12

This answer makes sense - there will be a high concentration of OH because it's basic, and a low concentration of H.

This was sort of an abstract way of doing it but it is how I always did it, easiest for me to remember.
Okay, the info given in the question is that [NaOH] = 2.38x10^-3 M

Since, NaOH dissociates completely in water, the concentration of OH- ions is also 2.38x10^-3 M.

We know that pOH = -lg([OH-]), so

pOH = -lg (2.38x10^-3) = 2.62342
(We use a lot of significant figures before we get the answer so as to reduce the error in our calculations)

Now, we know that pH = pKw - pOH where pKw = 14

So, pH=14-2.62342=11.37658

Now, we know that pH = -lg([H+]), so

-lg([H+]) = 11.37658
lg([H+]) = -11.37658
[H+] = 10^(-11.37658)
[H+] = 4.20x10^-12 M (Ans)

Hope this helps =)
1/2.38 x 10^-11 M

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