A50°c solution has a ph of 3.30. at this temperature, kw = 5.48 x 10 9. what is the poh of the solution?

The pOH of the solution is 10 .

Given that, the pH of the solution is 3.30, we also know that;

pH = - log[H^+]

Hence;

[H^+] = Antilog (-3.3)

[H^+] = 5.0 × 10^-4 M

But;

Kw = [H^+] [OH^-]

[OH^-] = Kw/[H^+]

The accurate value of Kw is 5.48 × 10^-14 not  5.48 x 10 9

[OH^-] = 5.48 x 10^-14/ 5.0 × 10^-4

[OH^-] = 1.1 x 10^-10 M

pOH = - log[1.1 x 10^-10 M]

pOH = 10

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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.

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