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What is the pH and pOH of a solution prepared by dissolving 1.40 g of Ca(OH)2 in water to make 865 mL of solution?

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User Reoxey
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First, we need to calculate the concentration of hydroxide ions (OH-) in the solution:

Number of moles of Ca(OH)2 = 1.40 g / 74.10 g/mol = 0.0189 mol
Volume of solution = 865 mL = 0.865 L

Concentration of OH- = 2 x 0.0189 mol / 0.865 L = 0.0437 M

From this, we can calculate the pOH:
pOH = -log[OH-] = -log(0.0437) = 1.36

Using the fact that pH + pOH = 14, we can calculate the pH:
pH = 14 - pOH = 14 - 1.36 = 12.64

So, the pH of the solution is 12.64 and the pOH is 1.36.

(If this doesn’t seem right to you make sure you comment!)
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User Liviu Chircu
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