No, I don't think so. First, pH = 8.75 gives (H^+) = 1.77 x 10^-9.
Then I would say (OH^-) = Kw/(H^+) = ??
Then BOH <==>B^+ + OH^-
Kb = (B^+)(OH^-)/(BOH) = 1.4 x 10^-11
You now have (B^+)=(OH^-) = 5.62 x 10^-6 and you can solve for (BOH).
Please let me know if my answer to the following question is correct.
Question:
What is the concentration of a weak base if its Kb = 1.4 x 10-11 and its pH = 8.75?
Answer:
Converting pH to H^+ concentration using the formula H^+ = 10^-ph, therefore 10^-8.75 gives an H^+ concentration value of 1.0 x 10^-9. Now, using the formula Ka=Kw/Kb and subsitituing in the values (Ka = 1.4 x10^-14/1.4 x 10^-11) gives 1.0 x 10^-3.
Now, subtracting 1.0 x 10^-3 from 1.0x 10^-9 gives a concentration of 9.9 x 10^-4 mol/L
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