[InsaneDefense]

Alright guys, it looks like I got what I was looking for.

From concentration of Calcium Hydroxide in the solution you antilog it.

pOH=-log0.0843 mol/L = 1.07417

pH = 14 - pOH - > 14 - 1.0742.

That was all. and to find the concentration I would use the negative of pH or pOH as a power of 10, so [OH^-]=10^-pOH or [H3O+]=10^-pH.


Thanks for the help though guys.

[AWK]

pOH=-log0.0843 mol/L = 1.07417

Wrong calculation.

[Borek]

From concentration of Calcium Hydroxide in the solution you antilog it.

pOH=-log0.0843 mol/L = 1.07417

No.

pH = 14 - pOH - > 14 - 1.0742.

That would be OK if not for the fact you ignored all advice given about the first step.

Think again: you have a 0.1M solution of CaCl₂. What is concentration of Ca²⁺? What is concentration of Cl^-? Is it in any general way different from the solution of Ca(OH)₂?