6) A 45 ml 0.09 M KOH solution is titrated with 0.11 M HCI: a) How...

Question

Question

Answer 1

Answer #1

a)

At equivalence point, number of moles of HCl will be equal to number of moles of KOH

HCl + KOH ----------- KCl + H2O

volume of KOH (in mL) * molarity of KOH = volume of HCl (in mL) * molarity of HCl

45 * 0.09 = V * 0.11

V = 36.82 mL

b) Half mid point of titration

Volume of HCl used = 36.82/2 = 18.41 mL

Total volume = 45 + 18.41 = 63.41 mL

Number of moles of KOH left = 1/2 * 45/1000 * 0.09 = 0.002025 moles

Molarity of KOH = 0.002025/63.41 * 1000 = 0.0319 M

pOH = -log[OH-] = -log[0.0319] = 1.49

pH = 14 - pOH = 14 - 1.49 = 12.51

c) pH at equivalence point will be equal to 7 since the base will be completely neutralized by the acid

Answer 2

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