11. How many grams of KClO 3 are required to make 50.0 mL of 0.100 M KClO 3 solu

Question

11. How many grams of KClO3 are required to make 50.0 mL of 0.100 M KClO3 solution? Write instructions on how the solution would be prepared.

13. Write a chemical equation for the ionization of each of the following in water AND indicate whether they are an acid, base, or salt. Use an equilibrium sign if it is weak.

a. Ba(OH)2

b. HNO2

c. NH3

14. Use the terms hypotonic, isotonic, or hypertonic to describe the solution in the following cases and answer the questions.

a.   A cucumber is soaked in some salt water (10% sodium chloride). The term to describe the

salt water (10% solute) is_______________________.

What happens to the cucumber cells in the salt water?

b.   A celery stalk is placed in pure water (0% NaCl). The term to describe the pure water (0% solute) is_________________________

What happens to the cells in the celery when they are placed in pure water?

c. An IV solution contains normal saline (a solution with 0.85%NaCl). The term for the normal saline is

________________________

What happens to the cells in the body when they are in contact with normal saline?

15.   Answer the question below regarding a buffer that is 0.5 M HNO2 and 0.5 M KNO2 .

a.What is the function of a buffer?

b.Which part of a HNO2 /KNO2 buffer reacts with added H+?Write a chemical equation for this reaction.

c.Which part of a HNO2 /KNO2 buffer reacts with added OH-1? Write a chemical equation for this reaction.

d.    What happens to proteins in the body if the solutions they are in are not buffered?

Explanation / Answer

11) Molar mass of KClO3 = 122.5 g/mol

Molarity = (mass/ molar mass) x (1/ volume of solution in litres)

Hence,

Molarity of KClO3 = (mass of KClO3 / molar mass of KClO3 ) x (1/ volume of solution in litres)

0.1 M = ( mass of KClO3 / 122.5 g/mol) x (1/ 0.05 L)

mass of KClO3 = 0.6125 g

Therefore,

mass of KClO3 required =0.6125 g

13) Ba(OH)2 -------------> Ba2+ + 2OH-

HNO2 + H2O <---------> H3O+ + NO2-

NH3 + H2O <-----------> NH4+ + OH-

15)

b) KNO2 + H+ -------------> HNO2 + K+

c) HNO2 + OH- -------------> NO2- + H2O

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