If a 24mL solution containing 2.0M HCl is mixed with a 40mL solution containing 0.25M Ca(OH)2. What is the pH of the resulting mixture? (pH = -log[H+]) Show ALL steps ON PAPER

Answer 1

Look at the reaction taking place:

2HCl + Ca(OH₎₂ ==> CaCl₂ + 2H₂O

Step 1: find moles of HCl present

24 ml x 1 L/1000 ml x 2.0 mol/L = 0.048 moles HCl

Step 2: find moles Ca(OH)₂ present:

40 ml x 1 L/1000 ml x 0.25 mol/L = 0.01 moles Ca(OH)₂

Step 3: Find which is in excess and by how much:

Since it takes 2 mol HCl for every 1 mol Ca(OH)₂, the HCl is present in excess. Here's by how much...

0.01 mol Ca(OH)₂ x 2 mol HCl/mol Ca(OH)₂ = 0.02 moles HCl used up

moles HCl left over = 0.048 mol - 0.02 mol = 0.028 mole HCl remaining

Step 3: Calculate concentration of HCl in final volume of 24 ml + 40 ml = 64 ml = 0.064 L:

0.028 mol / 0.064 L = 0.4375 M

Step 4: Find pH

pH = -log 0.4375

pH = 0.36