Calculate the pOH and pH at 298K of soloutions having the following ion concentrations

A.[OH-] = 1.0 X 10^-12
B.[OH-] = 1.3 X 10^-2

answered below.

To calculate the pOH and pH of solutions with given ion concentrations, we need to use the formulas related to the ion product of water (Kw) and the definitions of pOH and pH.

1. Calculation for solution A with [OH-] = 1.0 x 10^-12 M:
To calculate pOH, we use the formula:
pOH = -log[OH-]

Given [OH-] = 1.0 x 10^-12 M, we can substitute this value into the formula:
pOH = -log(1.0 x 10^-12)

Using the logarithmic property, we can simplify the calculation:
pOH = -log(1.0) - log(10^-12)
pOH = -log(1) + 12 log(10)

Since log(1) = 0, and log(10) = 1, the calculation becomes:
pOH = 12

To calculate pH, we use the relationship between pH and pOH:
pH + pOH = 14

Since we know pOH = 12, we can substitute this value into the equation:
pH + 12 = 14

Rearranging the equation to solve for pH:
pH = 14 - 12
pH = 2

Therefore, for solution A with [OH-] = 1.0 x 10^-12 M:
pOH = 12
pH = 2

2. Calculation for solution B with [OH-] = 1.3 x 10^-2 M:
Using the same formulas as above, we can calculate pOH and pH:

pOH = -log(1.3 x 10^-2)
pOH ≈ 1.89 (rounded to two decimal places)

pH + pOH = 14
pH + 1.89 = 14
pH ≈ 14 - 1.89
pH ≈ 12.11 (rounded to two decimal places)

Therefore, for solution B with [OH-] = 1.3 x 10^-2 M:
pOH ≈ 1.89
pH ≈ 12.11 (rounded to two decimal places)

Note: In these calculations, we assumed that the solutions are at 298K and that the concentration of H+ (or [H3O+]) in the solutions is negligible compared to the [OH-] concentrations given.