1.6 g of NaOH dissolves in water to a total volume of 300 mL. Calc. the molarity pOH and PH of this solution?

Answer 1

Molar mass of NaOH = 40 gmol-1

No of mol = mass / molar mass = 1.6 / 40 = 0.04 mol

0.04 mol in 300mL
? in 1000mL --> 0.04 * 1000 / 300 = 0.133 M

pOH = -log [OH-] = -log [0.133] = 0.88
pH = 14 - pOH = 13.12

Answer 2

Change grams to moles (divide by the molar mass of NaOH)

Molarity = moles/liters so divide your moles by the volume IN LITERS; this gives you the molarity of the NaOH.

Since NaOH is a strong base (dissociates 100%) the molarity of the OH will be with same as the molarity of the NaOH.

pOH = -log[OH] where [OH] = molarity of the OH

Since pH + pOH = 14, then pH = 14-pOH

Answer 3

Take 1.6 NaOH and convert to moles. 1.6g *(1mol/40g)=.04mol Then take that number and divide by .3L to get molarity answer. (.3L=300mL) Molarity comes out to be .13333M
Since NaOH is a strong acid, it will completely dissociate, thus we can assume that .13333M is the concentration of OH-
Now all you have to do is pOH=-log(OH)=.875
Since, pH+pOH=14, now take .875 and subtract it from 14, this is pH=13.1249.
You can use common knowledge to check answer, strong base is high pH, and that is what we got pH=13.1249