Question

Question #ad7e1

Answer 1

The [OH-] = 0.18 M
pOH = 0.74

[H+]= 5.6 x 10^-14

The questions asks you to connect several ideas:
1) calculation of molarity
The molar concentration of NaOH is calculated by dividing the mass of 8.6 grams by the molar mass of NaOH which is 40.0 g/mol.
This gives the number of moles which is 0.215 moles

This value is then divided by the volume in liters, because

                               molarity=  moles of [solute](http://socratic.org/chemistry/solutions-and-their-behavior/solute)/ liters of solution

                               Molarity = 0.215 moles/ 1.2 liters

                              Molarity of NaOH = 0.18 M

2) the second question to consider is the ionization of a strong base.
A strong base ionizes 100 %.

NaOH --> Na+ + OH-

So the concentration of OH- = concentration of NaOH

Therefore [OH-]= 0.18 M

3) you are also asked to understand the relationship between [OH-]and pOH.

To obtain the pOH from the [OH-] take the -log [OH-]

               pOH = -log [0.18]
                pOH = 0.74

4) then you are asked to relate pH and pOH
Both values equal 14.

pH = 14-pOH

pH = 14 - 0.74

pH. = 13.26

5) finally, you are asked to relate pH and [H+].

antilog (-pH) = [H+]

 antilog (- 13.25) = 5.6 x10 ^-14

Therefore, the hydrogen ion concentration is 5.6 x10^-14