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Answer
Given: conc. of HBr = 1.4 M
Volume of HBr = 15.4 mL
Volume of KOH = 22.10 mL

We know that, M1V1 = M2V2
                        (HBr)      (KOH)

Therefore, M2 = M1V1/V2
                        = 1.4 X 15.4/22.10
                        = 0.9756 M

Concentration of KOH is 0.9756 M.
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Given the data from the question, the concentration of the KOH solution obtained is 0.98 M

Balanced equation

HBr + KOH —> KBr + H₂O

From the balanced equation above,

  • The mole ratio of the acid, HBr (nA) = 1
  • The mole ratio of the base, KOH (nB) = 1

How to determine the molarity of KOH

  • Volume of acid, HCl (Va) = 15.4 mL
  • Molarity of acid, HCl (Ma) = 1.4 M
  • Volume of base, KOH (Vb) = 22.1 mL
  • Molarity of base, KOH (Mb) =?

MaVa / MbVb = nA / nB

(1.4 × 15.4) / (Mb × 22.1) = 1

21.56 / (Mb × 22.1) = 1

Cross multiply

Mb × 22.1 = 21.56

Divide both side 22.1

Mb = 21.56 / 22.1

Mb = 0.98 M

Thus, the concentration of the KOH solution needed is 0.98 M

Complete question

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