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