Chemistry help Acid and Bases:

Question

A solution is prepared by adding 1.00 × 10^-13 mole of strong base NaOH to 1.00 L of aqueous solution. The pH is:

A solution is prepared by adding 1.00 × 10^-2 mole of strong base NaOH to 1.00 L of aqueous solution. The pH is:

What is the differences in solving these two problems?

Answer 1

The difference in solving the two problems is that in the first situation the concentration of OH- ions contributed by the NaOH is so small that the OH- from H2O must be considered. In the 2nd situation, this is not the case.

1.00x10^-13 = 10^-13M OH-

OH- from H2O = 10^-7 M

Considering significant figures, tge pH would be 7.00

1.00x10^-2 M NaOH = 1.00x10^-2 M OH-

pOH = 2.00

pH = 12.00