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645 mL solution of HBr is titrated with 1.51 M KOH. If it takes 645.0 mL of the base solution to reach the equivalence point, what is the pH when only 125 mL of the base has been added to the solution?

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Answer:

pH = 13.38

Explanation:

  • HBr + KOH ↔ KBr + H2O

equivalence point:

  • (C×V)acid = (C×V)base

⇒ CHBr = ((1.51 M)(0.645 L))/(0.645 L ) = 1.51 M

pH after 125 mL of KOH added:

⇒ CHBr = ((1.51 M)(0.645 L) - (1.51 M)(0.125 L)) / (0.645 + 0.125)

⇒ CHBr = 1.02 M

  • HBr + H2O → H3O+  +  Br-
  • KOH + H2O → K+  + OH-

∴ CKOH = ((1.51 M)(0.125 L)) / (0.645 + 0.125)

⇒ CKOH = 0.245 M

∴ [OH-] = CKOH

⇒ pOH = - Log (0.245)

∴ pOH = 0.61

⇒ pH = 14 - pOH

⇒ pH = 13.38

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