Question

In a titration experiment a student uses 1.4 m hbr solution and the indicator phenolphthalein to determine the concentration of a koh solution.

Answer 1

The question is incomplete. Complete question is attached below:
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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.

Answer 2

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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