A 80.0 mL volume of 0.25 M HBr is titrated with 0.50M KOH. Calculate the pH after addition of 40.0 mL of KOH.

Answer 1

HBr + KOH ==> KBr + HOH

moles HBr initially = L x M = ?
moles KOH added = L x M = ?
See which is in excess and calculate pH from pH = -log (H^+). OR
if mole HBr = moles KOH, the the solution has just KBr and H2O and calculate pH from that. Post your work if you need additional assistance.

Answer 2

Find the moles of HBr = (liters)(M)
Find the moles of KOH = (liters(M)
If moles of HBr = moles KOH, pH = 7

If moles of HBr is larger than moles KOH, the mixture is acid and
[H+] = [(moles HBr)-(moles KOH)]/(total liters)
pH = -log[H+]

If moles of KOH is larger than moles of HBr,
[OH-] = [(moles KOH)-(moles HBr)]/(total liters)
pOH = -log[OH-]
and
pH = 14-pOH

Answer 3

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