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20.9: Worksheets for Chapter 20

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pH Worksheet

CK-12 Foundation Chemistry

Name______________________ Date_________

  1. Calculate the pH of a solution with [H^+] = 7.0 \times 10^{-5} \ M.
  2. Calculate the pH of a solution that is 0.050 \ M \ NaOH.
  3. Calculate the pH of a solution that is 7.0 \ M \ 10^{-5} \times Mg(OH)_2.
  4. What is the [H^+] in a solution with pH = 4.4?
  5. What is the [OH^-] in a solution with pH = 3.0?
  6. 10.0 \ g of KOH is added to enough water to make 400. \ mL of solution. What is the pH?
  7. A 1.0 \ liter solution has a pH = 2. How many liters of water must be added to change the pH to 3?
  8. If you do the regular calculations to determine the pH of a 1.0 \times 10^{-12} \ M \ HBr solution, you will get the pH = 12. You should have a feeling that something is wrong with this situation because this indicates that a solution of acid has a basic pH. What do you think is wrong with this calculation?

Complete the following table.

Acid, Base, or Neutral
pH [H^+] [OH^-] A, B, or N
9. 6.2 \times 10^{-4} \ M
10. 8.5 \times 10^{-10} \ M
11. 10.75
12. 4.0 \times 10^{-2} \ M

Strong Acids and Bases Worksheet

CK-12 Foundation Chemistry

Name______________________ Date_________

  1. If the hydrogen ion concentration in a solution is 1.00 \times 10^{-4} \ M, what is the hydroxide ion concentration?
  2. What is the hydroxide ion concentration in a solution whose pH is 11?
  3. What is the hydrogen ion concentration in a solution prepared by dissolving 0.400 \ grams of NaOH in enough water to make 2.00 \ liters of solution?
  4. How many mL of 0.100 \ M potassium hydroxide are required to neutralize 75.0 \ mL of 0.500 \ M \ HNO_3?
  5. If 50.0 \ mL of H_2SO_4 are neutralized by 100. \ mL of 0.200 \ M \ LiOH, what is the molarity of the H_2SO_4?
  6. What volume of 6.00 \ M \ HCl would be necessary to neutralize 400. \ mL of 3.00 \ M \ Ba(OH)_2?
  7. 200. \ mL of 0.0150 \ M \ NaOH is mixed with 300. \ mL of 0.00100 \ M \ HCl. What is the final [H^+] and [OH^-]?
  8. What is the pH of the final solution in problem 7?
  9. 700. \ mL of 1.00 \times 10^{-4} \ M \ H_2SO_4 is mixed with 300. \ mL of 1.00 \times 10^{-3} \ M \ Ba(OH)_2. What is the final [H^+] and [OH^-]?
  10. What is the pH of the final solution in problem 9?
  11. 25.0 \ mL of 0.0100 \ M \ HCl is mixed with 35.0 \ mL of 0.0300 \ M \ NaOH. What is the final [H^+] and [OH^-]?
  12. What is the pH of the final solution in problem 11?
  13. What is the final [H^+] and [OH^-] in a solution made by adding 100. \ mL of 0.000200 \ M \ HNO_3 to 100. \ mL of 0.0000990 \ M \ Ba(OH)_2?
  14. What is the pH of the final solution in problem 13?
  15. What is the molar mass of a solid monoprotic acid if 0.300 \ grams of the acid requires 30.0 \ mL of 0.200 \ M \ NaOH to neutralize it?

Weak Acids and Bases Worksheet

CK-12 Foundation Chemistry

Name______________________ Date_________

1. Explain the difference between the designations “strong” acid and “weak” acid.

2. The K_a of acid A is 6.4 \times 10^{-4} and the K_a of acid B is 1.7 \times 10^{-5}. Which acid is the stronger acid?

3. Explain what happens to the pH of a solution of acetic acid when a solution of sodium acetate is added to it.

4. Explain why a solution of sodium acetate will be basic.

5. What is the pH of a 0.0100 \ M solution of a weak acid, HX, if the K_a for HX is 8.1 \times 10^{-7}.

6. The pH of a 0.100 \ M solution of a weak acid, HQ, is 4.0. What is the K_a of this acid?

7. What is the pH of a 0.150 \ M solution of NH_4OH? The K_b for NH_4OH is 1.80 \times 10^{-5}.

8. The pH of a 1.00 \ M solution of the weak base methylamine is 12.3. The equation for the reaction of methylamine in water is

CH_3NH_{2(aq)} + H_2O \leftrightharpoons CH_3NH_{3(aq)}^+ + OH^-_{(aq)}.

What is the K_b for methylamine?

9. Will a 1.00 \ M solution of potassium acetate be acidic, basic, or neutral?

10. Will a 1.00 \ M solution of NH_4NO_2 be acidic, basic, or neutral? Use 1.8 \times 10^{-5} as the K_b for NH_4OH and 7.1 \times 10^{-4} as the K_a for HNO_2.

Conjugate Acids-Bases Worksheet

CK12 Foundation Chemistry

Name ________________________________ Date ______________

Conjugate Acid-Base Pairs

Acids and bases exist as conjugate acid-base pairs. The term conjugate come from the Latin meaning “joining together” and refers to things that are joined, particularly in pairs, such as Bronsted acids and bases.

Every time a Bronsted acid acts as a hydrogen ion donor, it forms a conjugate base. Imagine a generic acid, HA. When this acid donate a hydrogen to water, one product of the reaction is the A- ion, which is then capable of accepting a hydrogen ion, and is, therefore, a Bronsted base.

HA~~+~~H_2O~~ \leftrightharpoons ~~H_3O^+~~+~~A^-
 \text{ACID}~~~~~~~~~~~~~~~~~~~\text{BASE}

Conversely, every time a base gains an H^+, the product is a Bronsted acid.

A^-~~+~~H_2O~~ \leftrightharpoons ~~HA~~+~~OH^-
 \text{BASE}~~~~~~~~~~~~~~~~~~\text{ACID}

Acids and bases in the Bronsted model, therefore, exist as conjugate pairs whose formulas are related by the gain or loss of a hydrogen ion.

The use of the symbols HA and A^- for the conjugate acid-base pair does not mean that all acids are neutral molecules or that all bases are negative ions. It signified only that the acid contains a hydrogen ion that is not present in the conjugate base. Bronsted acids or bases can be neutral molecules, positive ions, or negative ions.

Exercises

1. Write the formula for the conjugate base of each of the following acids.

A. HCN . . . conjugate base = ______
B. HSO_4^- . . . conjugate base = ______
C. HF . . . conjugate base = ______
D. HNO_2 . . . conjugate base = ______

2. Write the formula for the conjugate acid of each of the following bases.

A. NH_3 . . . conjugate acid = ______
B. HCO_3^- . . . conjugate acid = ______
C. HS^- . . . conjugate acid = ______
D. Br^- . . . conjugate acid = ______

3. For each given formula, indicate whether its conjugate partner is an acid or base and write its formula.

A. SO_4^{2-} . . . conjugate ______ = ______
B. HI . . . conjugate ______ = ______
C. S^{2-} . . . conjugate ______ = ______
D. HNO_3 . . . conjugate ______ = ______

4. In each of the following acid-base reactions, identify the acid and base on the left and their conjugate partners on the right.

A. CHOOH_(aq)~~~+~~~H_2O_{(L)}~~~ \leftrightharpoons ~~~HCOO^-_{(aq)}~~~+~~~H_3O^+_{(aq)}
B. H_2S_{(aq)}~~~+~~~NH_{3(aq)}~~~ \leftrightharpoons ~~~NH^+_{4(aq)}~~~+~~~HS^-_{(aq)}

Answers to Worksheets

  • The worksheet answer keys are available upon request. Please send an email to teachers-requests@ck12.org to request the worksheet answer keys.

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Answers for pH Worksheet

1. Calculate the pH of a solution with [H^+] = 7.0 \times 10^{-5}\ M.

pH = -\log \ [H^+] = -\log \ (7.0 \times 10^{-5}) = 4.15

2. Calculate the pH of a solution that is 0.050 M NaOH.

\text{Since} \ [OH^-] & = 0.050\ M \ \text{then,}\ [H^+] = \frac {1.0 \times 10^{-14}} {0.050} = 2.0 \times 10^{-13}\ M\\pH & = -\log \ [H^+] = -\log \ (2.0 \times 10^{-13}) = 12.7

3. Calculate the pH of a solution that is 7.0 \times 10^{-5}\ M\ Mg(OH)_2.

\text{If} \ [Mg(OH)_2] & =  7.0  \times 10^{-5} \ M,\  \text{then}\  [OH^-] =  (2)(7.0 \times  10^{-5} \ M) =  1.4 \times  10^{-4}\ M\\[H^+] & =  \frac {1.0 \times 10^{-14}} {1.4 \times 10^{-4}}  =  7.1 \times  10^{-11}\ M\\pH  & =  -\log \ [H^+] =  -\log \ (7.1 \times  10^{-11}) =  10.1

4. What is the [H^+] in a solution with pH = 4.4?

[H^+] = 10^{-pH} = 10^{-4.4} = 3.98 \times 10^{-5}\ M

5. What is the [OH^-] in a solution with pH = 3.0?

If pH = 3.0, then [H^+] =  1.0 \times  10^{-3}\ M and [OH^-] =  \frac {1.0 \times 10^{-14}} {1.0 \times 10^{-3}}  =  1.0 \times  10^{-11}\ M

6. 10.0 g of KOH is added to enough water to make 400. mL of solution. What is the pH?

mols\ KOH & =  \frac {10.0\ g} {56.1\ g/mol} = 0.178\ mol\\[KOH] & =  \frac {0.178\ mol} {0.400\ L} = 0.446\ M\\[H^+] & =  \frac {1.0 \times 10^{-14}} {0.446}  =  2.2  \times  10^{-14}\ M\\pH & =  -\log \  [H^+] =  -\log \  (2.2 \times  10^{-14})  =  13.6

7. A 1.0 liter solution has a pH = 2. How many liters of water must be added to change the pH to 3?

At pH = 2, [H^+] =  1.0 \times 10^{-2}\ M

If [H^+] = 1.0 \times 10^{-2}\ M, then 1.0 liter of solution contains 0.010 mol [H^+].

At pH = 3, [H^+] =  1.0 \times  10^{-3}\ M

\text{Total} \ liters = \frac {mols} {M}  =  \frac {0.010\ mol} {0.0010\ mol/L} = 10.0\ liters

Beginning with 1.0 liter and ending with 10.0 liters would require an additional 9.0 liters.

8. If you do the regular calculations to determine the pH of a 1.0 \times 10^{-12}\ M \ HBr solution, you will get the pH = 12. You should have a feeling that something is wrong with this situation because this indicates that a solution of acid has a basic pH. What do you think is wrong with this calculation?

This is such a dilute solution of HBr that the contribution of [H^+] from the dissociation of water is the predominant source of [H^+]. The pH of this solution will be nearly 7 because the contribution of hydrogen ion from the HBr is not significant.

Complete the following table.

pH [H^+] [OH^-] A, B, or N
9. 3.2 6.2 \times 10^{-4}\ M 1.6 \times 10^{-11}\ M A
10. 4.9 1.2 \times 10^{-5}\ M 8.5 \times 10^{-10}\ M A
11. 10.75 5.6 \times 10^{-10}\ M 1.8 \times 10^{-5}\ M B
12. 12.6 2.5 \times 10^{-13}\ M 4.0 \times 10^{-2}\ M B

Answers for Strong Acids and Bases Worksheet

1. If the hydrogen ion concentration in a solution is 1.00 \times 10^{-4}\ M, what is the hydroxide ion concentration?

[H^+][OH^-] & = 1.00 \times 10^{-14}\\[OH^-] & = \frac {1.00 \times 10^{-14}} {[H^+]} = \frac {1.00 \times 10^{-14}} {1.00 \times 10^{-4}} = 1.00 \times 10^{-10}\ M

2. What is the hydroxide ion concentration in a solution whose pH is 11?

[H^+] & = 10^{-pH} = 1.00 \times 10^{-11}\ M\\[OH^-] & = \frac {1.00 \times 10^{-14}} {[H^+]} = \frac {1.00 \times 10^{-14}} {1.00 \times 10^{-11}} = 1.00 \times 10^{-3}\ M

3. What is the hydrogen ion concentration in a solution prepared by dissolving 0.400 grams of NaOH in enough water to make 2.00 liters of solution?

moles\ NaOH & = \frac {0.400\ grams} {40.0\ grams/mole} = 0.0100\ mole\\[OH^-] & = \frac {0.0100\ mole} {2.00\ liter} = 0.00500\ M\\[H^+] & = \frac {1.00 \times 10^{-14}} {0.00500\ M} = 2.00 \times 10^{-12}\ M

4. How many mL of 0.100 M potassium hydroxide are required to neutralize 75.0 mL of 0.500 M HNO_3?

At neutrality, moles of hydrogen ion are equal to moles of hydroxide ion.

moles \ \text{hydrogen ion} & = (M)(L) = (0.500\ M)(0.0750\ L) = 0.0375\ moles\\moles \ \text{hydroxide ion} & = (0.100\ M)(x\ L) = 0.0375\ moles\\x & = 0.375\ liters = 375\ mL

5. If 50.0 mL of H_2SO_4 are neutralized by 100. mL of 0.200 M LiOH, what is the molarity of the H_2SO_4?

moles \ OH^- & = (0.200\ M)(0.100\ L) = 0.0200\ moles\\moles\ H^+ & = (x\ M)(0.0500\ L)(2) = (0.100)(x)\\moles\ H^+ & = moles\ OH^-\\(0.100)(x) & = 0.0200\\x & = 0.200\ M

6. What volume of 6.00 M HCl would be necessary to neutralize 400. mL of 3.00 M Ba(OH)_2?

moles \ \text{hydrogen ion} & = moles \ \text{hydroxide ion}\\(6.00\ M)(x\ L) & = (3.00\ M)(0.400\ L)(2)\\x & = 0.400\ L

7. 200. mL of 0.0150 M NaOH is mixed with 300. mL of 0.00100 M HCl. What is the final [H^+] and [OH^-]?

moles\ H^+ & = (0.00100\ M)(0.300\ L) = 0.000300\ moles\\moles\ OH^- & = (0.0150\ M)(0.200\ L) = 0.00300\ moles\\\text{excess}\ moles\ of\ OH^- & = 0.00300\ moles - 0.000300\ moles = 0.00270\ moles\\[OH^-] & = \frac {0.00270\ moles} {0.500\ liters} = 0.00540\ M\\[H^+] & = \frac {1.00 \times 10^{-14}} {0.00540} = 1.85 \times 10^{-10}\ M

8. What is the pH of the final solution in problem 7?

pH = - \log \ [H^+] = - \log\ (1.85 \times 10^{-10}\ M) = 11.7

9. 700. mL of 1.00 \times 10^{-4}\ M\ H_2SO_4 is mixed with 300. mL of 1.00 \times 10^{-3}\ M\ Ba(OH)_2. What is the final [H^+] and [OH^-]?

moles\ H^+ & = (2)(1.00 \times 10^{-4}\ M)(0.700\ L) = 1.40 \times 10^{-4}\ moles\\moles\ OH^- & = (2)(1.00 \times 10^{-3}\ M)(0.300\ L) = 6.00 \times 10^{-4}\ moles\\\text{excess}\ moles\ of\ OH^- & = 6.00 \times 10^{-4}\ M - 1.40 \times 10^{-4}\ M = 4.60 \times 10^{-4}\ moles \\[OH^-] & = \frac {4.60 \times 10^{-4}\ moles} {1.00\ liter} = 4.60 \times 10^{-4}\ M\\[H^+] & = \frac {1.00 \times 10^{-14}} {4.60 \times 10^{-4}\ M} = 2.17 \times 10^{-11}\ M

10. What is the pH of the final solution in problem 9?

pH = - \log \ [H^+] = - \log\ (2.17 \times 10^{-11}\ M) = 10.7

11. 25.0 mL of 0.0100 M HCl is mixed with 35.0 mL of 0.0300 M NaOH. What is the final [H^+] and [OH^-]?

moles\ H^+ & = (0.0100\ M)(0.0250\ L) = 0.000250\ moles\\moles\ OH^- & = (0.0300\ M)(0.0350\ L) = 0.00105\ moles\\\text{excess}\ moles\ of\ OH^- & = 0.000800\ moles\\[OH^-] & = \frac {0.000800\ moles} {0.0600\ liter} = 0.133\ M\\[H^+] & = \frac {1.00 \times 10^{-14}} {0.133\ M} = 7.52 \times 10^{-14}\ M

12. What is the pH of the final solution in problem 11?

pH = - \log\ [H^+] = - \log \ (7.52 \times 10^{-14}\ M) = 13.1

13. What is the final [H^+] and [OH^-] in a solution made by adding 100. mL of 0.000200 M HNO_3 to 100. mL of 0.0000990 M Ba(OH)_2?

moles\ H^+ & = (0.000200\ M)(0.100\ L) = 0.0000200\ moles\\moles\ OH^- & = (2)(0.0000990\ M)(0.100\ L) = 0.0000198\ moles\\\text{excess}\ moles\ of\ H^+ & = 2.00 \times 10^{-7}\ moles\\[H^+] & = \frac {2.00 \times 10^{-7}\ moles} {0.200\ liter} = 1.00 \times 10^{-6}\ M\\[OH^-] & = \frac {1.00 \times 10^{-14}} {1.00 \times 10^{-6}\ M} = 1.00 \times 10^{-8}\ M

14. What is the pH of the final solution in problem 13?

pH = - \log \ [H^+] = - \log \ (1.00 \times 10^{-6}\ M) = 6.00

15. What is the molar mass of a solid monoprotic acid if 0.300 grams of the acid requires 30.0 mL of 0.200 M NaOH to neutralize it?

moles\ OH^- & = (0.200\ M)(0.0300\ L) = 0.00600\ moles\\\text{Therefore,}\ moles\ H^+ & = 0.00600\ moles\\molar\ mass & = \frac {grams} {moles} = \frac {0.300\ grams} {0.00600\ moles} = 50.0\ grams/mole

Answers for Weak Acids and Bases Worksheet

  • The worksheet answer keys are available upon request. Please send an email to teachers-requests@ck12.org to request the worksheet answer keys.

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Aug 18, 2012

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CK.SCI.ENG.TE.2.Chemistry.20.9

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