Respuesta :
Answer:
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
Answer:
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
