1) A student is asked to standardize a solution of barium hydroxide . He weighs

Question

1)

A student is asked to standardize a solution of barium hydroxide. He weighs out 1.10 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid).

It requires 28.0 mL of barium hydroxide to reach the endpoint.

A. What is the molarity of the barium hydroxide solution?  M

This barium hydroxide solution is then used to titrate an unknown solution of nitric acid.

B. If 27.5 mL of the barium hydroxide solution is required to neutralize 12.8 mL of nitric acid, what is the molarity of thenitric acid solution?  M

2)A student is asked to standardize a solution of barium hydroxide. He weighs out 1.10 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid).

It requires 24.6 mL of barium hydroxide to reach the endpoint.

A. What is the molarity of the barium hydroxide solution?  M

This barium hydroxide solution is then used to titrate an unknown solution of hydroiodic acid.

B. If 14.1 mL of the barium hydroxide solution is required to neutralize 30.0 mL of hydroiodic acid, what is the molarity of the hydroiodic acid solution?  M

Explanation / Answer

Solution:

Question .no : 1

2 KHP + Ba(OH)2 -----> Ba(KP)2 + 2H2O

a) moles of KHP = 1.10 g / 204.22 g/mol = 0.005386 moles of KHP

    0.005386 moles KHP (1 mole Ba(OH)2 / 2 moles KHP) = 0.002693 moles Ba(OH)2

    Molarity of Ba(OH)2 = 0.002693 / 0.028 lt = 0.09618 M

b) Ba(OH)2 + 2HNO3 ----> Ba(OH)2 + 2H2O

    moles of Ba (OH)2 = 0.09618 M * 0.0275 lt = 0.002645 moles of Ba(OH)2

    molesof HNO3 = 0.002645 moles Ba(OH)2 (2 moles HNO3 / 1 mole Ba(OH)2)

                          = 0.00529 moles

    Molarity of HNO3 = 0.00529 moles / 0.0128 lt = 0.41328 M

Question .no: 2

2 KHP + Ba(OH)2 -----> Ba(KP)2 + 2H2O

a) moles of KHP = 1.10 g / 204.22 g/mol = 0.005386 moles of KHP

    0.005386 moles KHP (1 mole Ba(OH)2 / 2 moles KHP) = 0.002693 moles Ba(OH)2

    Molarity of Ba(OH)2 = 0.002693 / 0.0246 lt = 0.1095 M

b) Ba(OH)2 + 2HI ----> BaI2 + 2H2O

    moles of Ba (OH)2 = 0.1095 M * 0.0141 lt = 0.001544 moles of Ba(OH)2

    molesof HI = 0.001544 moles Ba(OH)2 (2 moles HI / 1 mole Ba(OH)2)

                          = 0.003088 moles

    Molarity of HI = 0.003088 moles / 0.030 lt = 0.1029 M

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