Chemistry Chapter 19 Practice with acids and bases.

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Chemistry Chapter 19 Practice with acids and bases

Transcript of Chemistry Chapter 19 Practice with acids and bases.

Page 1: Chemistry Chapter 19 Practice with acids and bases.

Chemistry Chapter 19

Practice with acids and bases

Page 2: Chemistry Chapter 19 Practice with acids and bases.

Calculate the pH

• A solution is found to contain 3.65 x 10-3 mol/liter HCl, a strong acid.– What is the pH of the solution?

– What is the hydroxide ion concentration in the solution?

Page 3: Chemistry Chapter 19 Practice with acids and bases.

Calculate the pH

• A solution is found to contain 3.65 x 10-3 mol/liter HCl, a strong acid.– What is the pH of the solution?

» pH = -log[H+]

» - log(3.65 x 10-3) = 2.44

» Because the solution has an ion concentration of greater than 1 x 10-7, it is an acidic solution.

» Also because the pH is less than 7, we know it is acidic.

Page 4: Chemistry Chapter 19 Practice with acids and bases.

Calculate the pH

• A solution is found to contain 3.65 x 10-3 mol/liter HCl, a strong acid.– What is the hydroxide ion concentration in the

solution?» OH- ion concentration can be found by using Kw.

» Kw = [H+][OH-]

» 1 x 10-14 = (3.65 x 10-3)(x)

» 2.74 x 10-12

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Working backwards

• A solution has a pH of 9.3. Using a mathematical approach, show that the solution is acidic or basic.

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Solution

• Acids have an H+ concentration of 1 x 10-7 or greater. We will therefore calculate the concentration using pH = -log[H+].9.3 = -log(x) or antilog -9.3 = x

x = 3.17 x 10-10

The solution is basic because the H+ ion concentration is less than 1 x 10-7

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More practice

• What is the pH of a solution that contains 5.44 x 10 –5 M NaOH?

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Solution

• What is the pH of a solution that contains 5.44 x 10 –5 M NaOH?

Two ways to do this:

solve for the H+, or calculate pOH and then use pKw to get pH

Page 9: Chemistry Chapter 19 Practice with acids and bases.

Solution

• What is the pH of a solution that contains 5.44 x 10 –5 M NaOH?

• 1 x 10-14 = [H+][OH-]

• 1 x 10-14/5.44 x 10-5 = [H+] = 1.84 x 10-10

• pH = -log(1.84 x 10-10) = 9.7 - basic

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Other solution

• What is the pH of a solution that contains 5.44 x 10 –5 M NaOH?

• pOH = -log[OH-]

• -log(5.44 x 10-5) = 4.26

• pKw = pH + pOH or 14 – 4.26 = pH

• pH = 9.74 (answers are the same)

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Polyprotic acids

• Many acids release more than one proton.Examples: H2SO4, H3PO4, H2C2O4

When these dissociate, they change the pH differently. Strong acids will donate all protons from H+, weak will not donate even all the first ions.

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Example

• Calculate the pH of a 0.002 M H2SO4 solution.

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Solution

• Calculate the pH of a 0.002 M H2SO4 solution.

• [H+] = 2(0.002) 0r 0.004 M H+ ions

• pH = -log(0.004) = 2.40

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Problem

• What is the H+ ion concentration if the OH- ion concentration is 1 x 10-4 M? Kw for water is 1.0 x 10-14

• Is this solution an acid or a base?

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Problem

• Show the neutralization reaction of acetic acid, a weak acid, and sodium hydroxide, a strong base.

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Problem

• If I were to put sodium acetate salt into water, would the solution be acidic, basic, or neutral? Why?

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Problem

• What acids and bases will produce ammonium chloride?

• When placed in water, would the solution be acidic, basic, or neutral? Why?

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Problem

• When placed in water would the solution of sodium bromide be acidic, basic or neutral?

• What color would it turn a strip of litmus paper?

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Questions

• What consitutes a strong acid?

• What constitutes a weak acid?

• What is a conjugate acid? How does it differ from a conjugate base?

• What is the difference between Bronsted –lowry acids and Arrhenius acids?