How do you calculate the pOH of the aqueous solution that has a pH of 2.7 at 298K?

Answer 1

The ion-product of pure water is 2.51E-14 at 37 deg C. Because it is a amphoteric (neutral; both acidic and basic), the hydronium ion concentration (H3O+) and the hydroxide ion concentration are equal, and are therefore equal to the square root of the ion-product:

(H3O+)(OH-) = (x)(x) = x2 = ---> x = SQRT(2.51E-14) = (2.51E-14)1/2

The equation for pH = -log(H3O+) = -log(x) , so

for water at 37 deg C: pH = -log(2.51E-14)1/2 = 1/2*[-log(2.51E-14)] = 1/2*13.6 = 6.8

Answer 2

pH + pOH = 14

So if pOH=2.7,

2.7 + pH = 14

pH = 14-2.7

pH = 11.3

Answer 3

pOH=14-pH of 8.7=5.3 therefore pOH=5.3

Answer 4

.230 M

Answer 5

10.3