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# 20.9: Worksheets for Chapter 20

Difficulty Level: At Grade Created by: CK-12

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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. $H_2S_{(aq)}~~~+~~~NH_{3(aq)}~~~ \leftrightharpoons ~~~NH^+_{4(aq)}~~~+~~~HS^-_{(aq)}$

• 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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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.

Aug 18, 2012

Aug 13, 2014