Find the [OH^-1] if the pH is 11.7
A) 1.0 x 10^-12 M
B) 2.9 x 10^-11 M
C) 1.5 x 10^-12 M
D) 2.0 x 10^12 M

None of the options are correct.

The answer to the question is 5.01x10^-3M

Explanation:

The following were obtained from the question:

pH = 11.7

Recall: pH + pOH = 14

11.7 + pOH = 14

pOH = 14 — 11.7

pOH = 2.3

Now we can obtain the [OH^-1] as follows:

pOH = —Log [OH^-1]

pOH = 2.3

2.3 = —Log [OH^-1]

—2.3 = Log [OH^-1]

Take the anti-log of —2.3

[OH^-1] = 5.01x10^-3M

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.