Final answer:
To determine the pH of a solution created by dissolving 1.1 g of Ca(OH)2 in 950 mL of water, we calculate the moles of Ca(OH)2, determine its molarity, calculate the OH- concentration, and then calculate the pH, which is approximately 12.5.
Step-by-step explanation:
The question asks for the pH of a solution prepared by dissolving 1.1 g of Ca(OH)2 in water to make a 950 mL solution. To answer this, we must first calculate the molarity of the solution since pH is related to the hydrogen ion concentration. Calcium hydroxide is a strong base that dissociates fully in water to produce two hydroxide ions for every one calcium ion. First, calculate the moles of Ca(OH)2:
Molar mass of Ca(OH)2 = 40.08 (Ca) + (16.00 x 2) (O) + (1.008 x 2) (H)
= 74.096 g/mol
Moles of Ca(OH)2 = mass/molar mass
= 1.1 g / 74.096 g/mol
= 0.01484 moles.
Molarity (M) = moles/volume
= 0.01484 moles / 0.95 L
= 0.01562 M.
Since Ca(OH)2 produces two OH− ions,
OH− concentration = 2 x molarity of Ca(OH)2
= 2 x 0.01562 M
= 0.03124 M.
Next, calculate pOH:
pOH = -log[OH−]
= -log(0.03124)
= 1.505.
Finally, calculate pH using the relation between pH and pOH where pH + pOH = 14:
pH = 14 - pOH
= 14 - 1.505
= 12.495.
Therefore, the pH of the solution is approximately 12.5.