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 of solution is: 12.82

Explanation:

The reaction of hydrochloric acid with calcium hydroxide is:

Ca (OH) ₂ + 2 HCl → H₂O + CaCl₂

Where 2 moles of hydrochloric acid react with 1 mole of calcium hydroxide

Moles of HCl and moles of Ca (OH) ₂ are:

Moles HCl:

0.02500L * (0.155mol / L) = 3.875x10⁻³moles HCl

Moles of Ca (OH) ₂:

0.03500L * (0.112mol / L) = 3.92x10⁻³moles Ca (OH) ₂

Moles of Ca (OH) ₂ that react are:

3.875x10⁻³moles HCl * (1 mol Ca (OH) ₂ / 2 mol HCl) = 1.9375x10⁻³ mol Ca (OH) ₂

Thus, moles of Ca (OH) ₂ that remain are:

3.92x10⁻³moles Ca (OH) ₂ - 1.9375x10⁻³ mol Ca (OH) ₂ = 1.9825x10⁻³ mol Ca (OH) ₂

Moles of OH⁻ are:

1.9825x10⁻³ mol Ca (OH) ₂ * (2mol OH⁻ / 1mol Ca (OH) ₂) = 3.965x10⁻³ mol OH⁻

As volume is 25mL + 35mL = 60mL ≡ 0.060L. Molar concentration of OH⁻ is:

3.965x10⁻³ mol OH⁻ / 0.060L = 0.066M OH⁻.

pOH = - log [OH⁻]

pOH = 1.18

pH = 14-pOH

pH of solution is: 12.82

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