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?

pH = 13.38

Explanation:

equivalence point:

⇒ 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

∴ 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

beta decay is the answer

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