posted by Anonymous on .
What will be the pH of:
(a)10.0cm3 of an aqueous solution of 0.001mol dm^-3 nitric aicd, HNO3 (aq)?
(b)100cm3 of an aqueous solution of 0.001mol dm^-3 nitric aicd, HNO3 (aq)?
(c)0.02mol dm^-3 of potassium hydroxide solution, KOH (aq)?
For (a) I got 2, (b) 1 and (c) 12.31.
I am not sure if these are right. I used pH=-log[H+]/pH-log[OH-]. How is the volume used?
Thank you in advance for any help!
c is correct except I obtained 12.3.
I THINK the mistake you are making is you are getting confused with the volume.
a. It doesn't matter how much of the 0.001 M HNO3 you have, it still is 0.001 M; therefore, the pH = - log(H) = -log (10^-3) = 3.
b. same concn; therefore, same pH.
c. 0.02 M KOH means 0.02 M OH.
pOH = -log(0.02) = -(-1.699)= 1.699 = 1.7 and 14-1.7=12.3