Chemistry HW Help

Question

Calculate the pH when 56.0 mL of 0.250 M HCI is mixed with 40.0 mL of 0.150 M Ca(OH)2.

Answer 1

2HCl + Ca(OH)₂ ==> CaCl₂ + 2H₂O .. balanced equation

moles HCl present = 56.0 ml x 1 L / 1000 ml x 0.250 mol / L = 0.0140 mols HCl

moles Ca(OH)₂ present = 40.0 ml x 1 L / 1000 ml x 0.150 mol / L = 0.00600 mols Ca(OH)₂

It takes 2 mols HCl for each 1 mol Ca(OH)2 according to balanced equation. 2 x 0.006 mols = 0.0120 mols of HCl would be needed to react with all the Ca(OH)2. So, there will be 0.002 mols HCl left over. You can see this if you set up an ICE table:

2HCl + Ca(OH)₂ ==> CaCl₂ + 2H₂O

0.014.........0.006.............0...............Initial

-0.012.......-0.006..........+0.006........Change

0.002.............0................0.006........Equilibrium

The final volume = 56.0 ml + 40.0 ml = 96.0 ml = 0.096 L

Final concentration of HCl = 0.002 mols / 0.096 L = 0.0208 M

pH = -log [H+]

pH = -log 0.0208

pH = 1.68