What is the pH of 0.11 M solution of Ca(OH)2 [log 11 = 1.041]Select an answerA0.65     B0.95C2D0.3010

The pH of a solution can be calculated using the formula:

pH = -log[H+]

However, in the case of Ca(OH)2, it is a base and not an acid. So, we need to find the pOH first and then subtract from 14 to get the pH (since pH + pOH = 14).

First, we need to find the concentration of OH- ions. Each formula unit of Ca(OH)2 produces 2 OH- ions. So, the molarity of OH- ions is 2 * 0.11 M = 0.22 M.

Next, we find the pOH. pOH = -log[OH-] = -log(0.22).

We know that log 11 = 1.041. So, we can estimate that log 2.2 is approximately 0.342 (since 2.2 is 1/5 of 11, and 1.041/5 = 0.2082, which we subtract from 1 since the log of a number between 1 and 10 is between 0 and 1).

So, pOH = -(-0.342) = 0.342.

Finally, we find the pH by subtracting the pOH from 14.

pH = 14 - pOH = 14 - 0.342 = 13.658.

None of the given options (A0.65, B0.95, C2, D0.3010) match this result. There might be a mistake in the question or the provided options.