Question

1. An aqueous solution of potassium hydroxide is standardized by titration with a 0.117 M solution of hydrobromic acid. If 18.7 mL of base are required to neutralize 15.7 mL of the acid, what is the molarity of the potassium hydroxide solution?

2. .An aqueous solution of nitric acid is standardized by titration with a 0.183 M solution of calcium hydroxide. If 28.5 mL of base are required to neutralize 21.4 mL of the acid, what is the molarity of the nitric acid solution?

3.In the laboratory you dissolve 19.8 g of aluminum acetate in a volumetric flask and add water to a total volume of 125 mL.

What is the molarity of the solution?

M. What is the concentration of the aluminum cation? M.

What is the concentration of the acetate anion? M.

4. In the laboratory you dissolve 22.3 g of iron(II) chloride in a volumetric flask and add water to a total volume of 375 mL.

What is the molarity of the solution? M.

What is the concentration of the iron(II) cation? M.

What is the concentration of the chloride anion?

5. A 14.9 g sample of an aqueous solution of hydrobromic acid contains an unknown amount of the acid.
If 13.5 mL of 0.583 M sodium hydroxide are required to neutralize the hydrobromic acid, what is the percent by mass of hydrobromic acid in the mixture?

6.A 14.0 g sample of an aqueous solution of nitric acid contains an unknown amount of the acid.
If 26.6 mL of 0.725 M barium hydroxide are required to neutralize the nitric acid, what is the percent by mass of nitric acid in the mixture?

Answer 1

Answer 1-

Since HBr and KOH both are mono equivalent

M1V1(HBr)=M2V2(KOH)

M1 x 18.7 =0.117 x 15.7

M1 =0.098 M

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Answer-2

Since Ca(OH)2 is diequivalent and HNO3 is mono

So , 2×M1V1 (HNO3) =M2V2 (Ca(OH)2)

2×21.4xM1 = 0.183 x 28.5

M1 =0.122 M

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Answer-3

Aluminum acetate is Al(CH3COOH)3

Molarity = (no of moles of aluminum acetate)/volume in litre

=(19.8 x 1000)/(125x204.11)

=0.776 M

So molarity of cation = 0.776 M

Molarity of acetate anion = 3 x 0.776

=2.328 M

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Anwer -4

Iron (II)chloride means FeCl2

Molarity =( 22.3 x1000/(126.75x375)

=0.469 M

So molarity of Fe2+ = 0.469 M

Molarity of cl- =0.938 M

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