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

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error

where are the choice options?