R-0.0821 L atm/mol K 1. A solution prepared by mixing 50 grams of sodium hydroxi

Question

R-0.0821 L atm/mol K 1. A solution prepared by mixing 50 grams of sodium hydroxide into a total solution volume of 500 ml, what is the Molarity, % (w/v), and ppm of the solution? 2. Define solute and solvent. Explain the difference between a solution, and an emulsion by giving an example of each and describing how they appear different 3. Explain how breathing occurs using Boyle's law. Include and define volume and pressure in your answer. 3. What are the numerical values for the pH scale of áqueous systems? What equation relates hydronium amount to pH? What equation relates hydronium amount to hydroxide amount? Which is more acidic a pH of 3 or a pH of 1? How many times more acidic? List the six strong acids, formulas and names. List the four strong bases, formulas and names. 4. The HyO concentration of a solution of NaOH is 1 x 10-12, what is the pH? What is the OH-1 concentration?

Explanation / Answer

1. NaOH mass = 50 g

molarity of NaOH solution = 50 g/40 g/mol x 0.5 L = 2.5 M

concentration in ppm = 50,000 mg/0.5 L = 1.0 x 10^5 ppm

%w/v = 50 g x 100/500 ml = 10%

2. A small amount of solute is dissolved in a large amount of solvent.

Solution is a homogeneous mixture of solute and solvent together. For example sugar in water.

Emulsion is dispersion of solute into solvent in the form of minute droplets. For example milk is an emulsion of fat in water.

3. pH of aqueous solution = 14

Kw = 1 x 10^-14

[H3O+] (hydronium ion) and pH

pH = -log[H3O+]

Kw = [H3O+][OH-]

A solution of pH 1 would be more acidic than pH 3 soluiton by three orders of magniture

[H3O+] in pH 1 = 1 x 10^-1 M

[H3O+] in pH 3 = 1 x 10^-3 M

Strong acids

HNO3 nitric acid

H2SO4 sulfuric acid

HCl hydrochloric acid

HI hydroiodic acid

HBr hydrobromic acid

HClO4 perchloric acid

Strong bases

NaOH sodium hydroxide

KOH potassium hydroxide

Sr(OH)2 strongtium hydroxide

Ba(OH)2 barium hydroxide

Ca(OH)2 calium hydroxide

LiOH lithium hydroxide

4. given,

[H3O+] = 1 x 10^-12 M

pH = -log(1 x 10^-12) = 12

[OH-] = 1 x 10^-14/1 x 10^-12 = 1 x 10^-2 M

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