When a 17.7 mL sample of a 0.368 M aqueous hypochlorous acid solution is titrated with a 0.301 M aqueous barium hydroxide solution, what is the pH after 16.2 mL of barium hydroxide have been added

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

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