Question

A 50°C solution has a pH of 3.30. At this temperature, Kw = 5.48 x 10 9. What is the pOH of the solution?

Answer 1

Answer:

4.8

Explanation:

The equation relating pH and pOH at any temperature is pH+pOH=−logKw.

Therefore,

Kw=5.48×10−14

pH=−logKw−pOH=−log(5.48×10−14)−8.5=4.8

Answer 2

Answer:

9.93

Explanation:

Your value for Kw is incorrect. The correct value is 5.48 × 10^-14.

  pH + pOH = pKw

3.30 + pOH = -log(5.84 × 10^-14) = 13.23

          pOH = 13.23 - 3.30 = 9.93

The pOH of the solution is 9.93.