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

Learn more: link

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.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.