Review Unit 6

Double Edged Sword

Vocab practice

• What is a hydrogen ion?– H+

• What is a hydroxide ion? -OH-

• What is a hydronium ion? -H3O+

What would be the pH of• Strong Acid

0• Strong base

14• Neutral

7• Concentrated weak acid

3• Dilute weak acid

6• Concentrated weak base

12• Dilute weak base

8

1. Which side shows an acid and which shows a base? How can you tell?

1. Which beaker shows a neutral solution? 2. Which beaker shows an acidic solution? 3. Which beaker shows a basic solution?

A B C

1. What is the strongest acid named in the diagram? 2. What is the strongest base named in the diagram?3. What is the weakest acid named in the diagram? 4. What is the weakest base named in the diagram?

1. Which is the strong acid? Weak acid? 2. Which would have the higher pH?

Identify the following images as concentrated/dilute and strong/weak acid.

A B C D

Draw the difference

• What is the difference between an acid and a base?

• What is the difference between diluted and concentrated?

• What is the difference between a strong acid and a weak acid?

Molarity Practice

• 1. What is the molarity of a solution containing 9.478 grams of RuCl3 in enough water to make 1.00L of solution? – Grams moles: 9.478 * (1 mole / 207.42 g)• 0.05 moles

– M = mol / L• M = 0.05 mol / 1.00 L• M = 0.05M

• 2. What is the molarity of a solution containing 5.035 grams of FeCl3 in enough water to make 500mL of solution?– Grams mol: 5.035 g * (1 mol/162.2g)• 0.03 mol

– M = mol / L• M = 0.03mol / .500L = • M = .06M

• 3. What is the molarity of a solution containing 72.9 grams of HCl in enough water to make 500mL of solution?– Grams mol: 72.9g * (1 mol / 36.46g)• 2 mol

– M = mol / L• M = 2 mol / .500L• M = 4M

Dilutions Practice

• 1. If I have 340mL of a 0.5M NaBr solution, what will be the concentration if I add 560mL more water to it?– M1V1 = M2V2

– 0.5 * 340 = M * 900– M = .19M

• 2. If I dilute 250mL of 0.10M lithium acetate solution to a volume of 750mL, what will the concentration be?– M1V1 = M2V2

– 0.10 * 250 = M * 750– M = .03M

• 3. If I leave 750mL of 0.50M sodium chloride solution uncovered on a windowsill and 150mL of the solvent evaporates, what will be the new concentration of the sodium chloride solution be?– M1V1 = M2V2

– 0.50 * 750 = M * 600– M = .625M

pH

• 1. What is the pH of a solution that contains 25 grams of hydrochloric acid (HCl) dissolved in 1.5 liters of water?– Grams mol: 25g * (1mol / 36.46g)

• 0.69mol

– M = mol / L• M = 0.69mol / 1.5L• M = 0.46M

– pH = -log [concentration of H+]• pH = -log [0.46]• pH = 0.34

• 2. What is the pH of a solution that contains 1.32 grams of nitric acid (HNO3) dissolved in 750mL of water?– Grams mol: 1.32g * (1mol /63.02g)

• 0.02mol

– M = mol / L• M = 0.02mol /.750L• M = 0.03M

– pH = -log [concentration of H+]• pH = -log [0.03]• pH = 1.52

• 3. What is the pH of a solution that contains 1.2 moles of nitric acid (HNO3) and 1.7 moles of hydrochloric acid (HCl) dissolved in 1000 liters of water?– M = mol / L

• M = (1.2mol + 1.7mol) / 1000L• M = 2.9mol / 1000L• M = .0029

– pH = -log[molarity]• pH = -log (.0029)• pH = 2.54

Challenge

• 6. How many grams of NaCl are required to prepare 100mL of a solution of 1M NaCl? – M = mol / L– Rearrange to solve for mol:• mol = M * L• mol = 1M * .100L• mol = 0.1mol

– Mol grams• 0.1mol * (58.44 g / 1mol)• 5.844 grams

• 4. What is the molarity of a solution containing 11.522 grams of KOH in enough water to make 350mL of solution?– Grams mol: 11.522g * (1 mol / 56.11g)• 0.21mol

– M = mol / L• M = 0.21 mol / .350L• M = .6M

• 5. What is the molarity of a solution containing 72.06 grams of BaCl2 in enough water to make 800mL of solution?– Grams mol: 72.06g * (1 mol / 208.23g)• 0.35mol

– M = mol / L• M = 0.35mol / .800L• M = 0.44M

• 7. How many grams of KMnO4 are required to prepare 1.0L of a solution of 1.5M KMnO4?– M = mol / L– Rearrange to solve for mol:

• mol = M * L• mol = 1.5M * 1.0L• mol = 1.5 mol

– Mol grams• 1.5mol * (158.04 g / 1mol)• 237.06 grams

• 8. How many grams of HNO3 are required to prepare 500mL of a 0.601M HNO3 solution?– M = mol / L– Rearrange to solve for mol:

• mol = M * L• mol = 0.601M * .500L• mol = 0.30 mol

– Mol grams• 0.30mol * (63.02 g / 1mol)• 18.91 grams

• 9. What is the volume of 0.1M HCl solution 1.46 grams of HCl?– Grams mol• 1.46 g * (1mol / 36.46g) • 0.04mol

– M = mol / L– Rearrange to solve for liters (volume)• L = mol / M• L = 0.04mol / 0.1M• L = 0.4L

• 10. What is the volume of a 0.2M AgNO3 solution containing 8.5 grams of AgNO3?– Grams mol

• 8.5 g * (1mol / 169.88g) • 0.05mol

– M = mol / L– Rearrange to solve for liters (volume)

• L = mol / M• L =0.05 / 0.2M• L = 0.25 L

• 4. To what volume would I need to add water to the evaporated solution in problem 3 to get a solution with a concentration of 0.25M?– M1V1 = M2V2

– 0.25 * V = .625 * 600– V = 1500 L – So need to add 900 more liters

• 4. If a solution has a [H+] concentration of 4.5 x 10-7 M, is this an acidic or basic solution? Explain.– pH = - log [concentration of H+]– pH = -log (4.5E-7)– pH = 6.35– That means that the solution is a weak acid

because the pH is close to neutral, but under 7

• 5. An acidic solution has a pH of 4. If I dilute 10mL of this solution to a final volume of 1000mL, what is the pH of the resulting solution?– pH of 4: use 10-pH to find concentration

• 10-4 = .0001 M

– M1V1 = M2V2

• .0001M * 10 = M * 1000• M = .000 001M

– pH = -log[M]• pH = -log(.000 001)• pH = 6