Calculate the pH of a solution prepared by mixing 25.00 mL of 0.155 M Ca(OH)2

solution with 35.00 mL of 0.112 M HCl solution

pH = 12.80

Explanation:

Step 1: data given

Volume of a 0.155 M Ca(OH)2 = 25.00 mL = 0.025 L

Volume of a 0.112 M HCl = 35.00 mL = 0.035 L

Step 2: The balanced equation

Ca(OH)2 + 2HCl → CaCl2 + 2H2O

Step 3: Calculate moles

Moles = molarity * volume

Moles Ca(OH)2 = 0.155M * 0.025 L

Moles Ca(OH)2 = 0.003875 moles

Moles HCl = 0.112 M * 0.035 L

Moles HCl = 0.00392 moles

Step 4: Calculate the limiting reactant

For 1 mol Ca(OH)2 we need 2 moles HCl to produce 1 mol CaCl2 and 2 moles H2O

HCl is the limiting reactant. It will completely be consumed. (0.00392 moles) Ca(OH)2 is in excess. There will react 0.00392 / 2 = 0.00196 moles

There will remain 0.003875 - 0.00196 = 0.001915 moles Ca(OH)2

Step 5: Calculate molarity of Ca(OH)2

Molarity = moles / volume

Molarity Ca(OH)2 = 0.001915 moles / (0.060 L)

Molarity Ca(OH)2 = 0.0319 M

For 1 mol Ca(OH)2 we have 2 moles OH-

Molarity of OH- = 2*0.0319 = 0.0638 M

Step 6: Calculate pH

Since Ca(OH)2 is a strong base

The ph will be the pH of a strong base

pOH = -log[0.0638)

pOH = 1.20

pH = 14 -1.20 = 12.80

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