Answer:
pH = 12.98
Explanation:
Step 1: Data given
Volume of aqueous hypochlorous acid solution = 17.7 mL = 0.0177 L
Molarity of aqueous hypochlorous acid solution = 0.368 M
Molarity of aqueous barium hydroxide solution = 0.301 M
Volume of aqueous barium hydroxide solution = 16.2 mL = 0.0162 L
Step 2: The balanced equation
2HCl + Ba(OH)2 → BaCl2 + 2H2O
Step 3: Calculate moles
Moles = molarity * volume
Moles HCl = 0368 M * 0.0177 L
Moles HCl = 0.0065136 moles
Moles Ba(OH)2 = 0.301 M * 0.0162 L
Moles Ba(OH)2 = 0.0048762 moles
Step 4: Calculate the limiting reactant
For 2 moles HCl we need 1 mol Ba(OH)2 to produce 1 mol BaCl2 and 2 moles H2O
HCl is the limiting reactant. It will completely be consumed 0.0065136 moles. Ba(OH)2 is in excess. There will react 0.0065136/2 = 0.0032568 moles. There will remain 0.0048762 moles - 0.0032568 = 0.0016194 moles
Step 5: Calculate molarity Ba(OH)2
Molarity Ba(OH)2 = moles / volume
Molarity Ba(OH)2 = 0.0016194 moles / 0.0339 L
Molarity Ba(OH)2 = 0.04777 M
Step 6: Calculate [OH-]
Ba(OH)2 → Ba^2+ + 2OH-
For Ba(OH)2 we have 2* [OH-]
[OH-] = 2*0.04777 = 0.09554 M
Step 7: Calculate pOH
pOH = -log[OH-]
pOH = -log(0.09554)
pOH = 1.02
Step 8: Calculate pH
pH = 14 - 1.02
pH = 12.98
