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What is the pH after 0.023 moles of HCl is added to a 100.0mL buffer solution...

What is the pH after 0.023 moles of HCl is added to a 100.0mL buffer solution containing 0.65M HOBr and 0.55M NaOBr ? Ka HOBr = 2.5 x 10-9

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Answer #1

Ka = 2.5 *10-9

pKa = -log ( 2.5 *10-9) = 8.60

Moles of HOBr = Molarity * Volume = 0.65 M *100 mL = 0.65 mol/L *0.100L = 0.065 moles

Moles of NaOBr = 0.55mol/L *0.100L = 0.055 moles

Now HCl is added, then it will reacts with NaOBr and forms HOBr

Thus, Moles of NaOBr added = 0.055 moles - 0.023 moles = 0.032 moles

Moles of HOBr = 0.065 moles + 0.023 moles = 0.088 moles

Using henderson Hasselbalch equation

pH = pKa + log[salt]/[acid]

pH = 8.60 + log( 0.032 /  0.088 ) = 8.16

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