Question

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?

Answer 1

pH = 13.38

Step-by-step 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