1. As I understand it--As an acid is added to neutral water the (OH-) will...decrease?? Is this right?

Question

2. If you made up a solution of NaOH by adding 0.010 mole of solid NaOH to 1.0 L of distilled water, what would be the concentration of the OH-?...I think this would be a pOH of 2, then I calculated the OH- to be 10^-2 which= .01.....Is this correct??

3. One mole of the acid HCL requires how many moles of OH- for neutralization? I think it is 1 mole...Is this correct? :)

4. A solution has a pH of 2.5. What is the (H+)?...I calculated it to be .00316 or 3.6 x 10^3? Is this right?
Thank-YOU so much!!!

Answer 1

Yes, yes, yes, yes.
However, you made a lot of extra work of 2.
If NaOH is 0.01 mol in 1 L, then OH^- is 0.01 M. You don't need to go through pOH and back to OH.