Question #ad7e1
1 Answer
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