Chapter 10 Acids and Bases

1. Chapter 10 Acids and Bases 10.10 Dilutions 10.11 Acid-Base Titration

2. Dilution Diluting a solution • Is the addition of water. • Decreases concentration. Concentrated Diluted Solution Solution

3. Diluting a Solution • In a dilution, the amount (moles or grams) of solute does not change. Amount solute (initial) = Amount solute (dilute) • A dilution increases the volume of the solution. Amount soluteAmount solute Volume (initial) Volume (increased) • As a result, the concentration of the solution decreases.

4. Calculations with Dilutions What is the molarity of an HCl solution prepared by adding 0.50 L of water to 0.10 L of a 12 M HCl. 1. Calculate the moles of HCl. 0.10 L x 12 moles HCl = 1.2 moles HCl 1 L 2. Determine the volume after dilution. 0.10 L + 0.50 L = 0.60 L solution 3. Calculate the molarity of the diluted HCl. 1.2 moles HCl = 2.0 M HCl 0.60 L

5. Dilution Equation • The dilution calculation can be expressed as an equation. C1V1 = C1V2 C = concentration • For molar concentration, the equation is M1V1 = M2V2 (moles) = (moles after dilution) • For percent concentration, %1V1 = %2V2 (grams) = (grams after dilution)

6. Calculation How many milliliters of 6.0 M NaOH solution are needed to prepare 1.0 L of a 0.15 M NaOH solution? M1 = 6.0 M M2 = 1.5 M V1 = ??? V2 = 1.0 L Rearrange the dilution equation for V1 V1 = M2V2 = 0.15 M x 1.0 L M1 6.0 M = 0.025 L x 1000 mL = 25 mL 1 L

7. Learning Check What volume (mL) of a 3.0 M H2SO4 solution can be prepared from 15 mL of 18 M H2SO4? 1) 90. mL 2) 27 mL 3) 15 mL

8. Solution 1) 90. mL M1 = 18.0 M M2 = 3.0 M V1 = 15 mL (0.015 L) V2 = ??? V2 = M1V1 = 18 M x 0.015 L = 0.090 L M2 3.0 M 0.090 L x 1000 mL = 90. mL 1L

9. Acid-Base Titration • Titration is a laboratory procedure used to determine the molarity of an acid. • In a titration, a base such as NaOH is added to a specific volume of an acid. Base (NaOH) Acid solution

10. Indicator • A few drops of an indicator is added to the acid in the flask. • The indicator changes color when the base (NaOH) has neutralized the acid.

11. End Point of Titration • At the end point, the indicator has a permanent color. • The volume of the base used to reach the end point is measured. • The molarity of the acid is calculated using the neutralization equation for the reaction.

12. Calculating Molarity What is the molarity of an HCl solution if 18.0 mL of a 0.25 M NaOH are required to neutralize 10.0 mL of the acid? 1. Write the neutralization equation. HCl + NaOH NaCl + H2O 2. Calculate the moles of NaOH. 18.0 mL NaOH x 1 L x 0.25 mole NaOH 1000 mL1 L = 0.0045 moles NaOH

13. Calculating Molarity (continued) 3. Calculate the moles of HCl. HCl + NaOH NaCl + H2O 0.0045 mole NaOH x 1 mole HCl 1 mole NaOH = 0.0045 moles HCl 4. Calculate the molarity of HCl. 10.0 mL HCl = 0.010 L HCl 0.0045 mole HCl = 0.45 M HCl 0.0100 L HCl

14. Learning Check Calculate the mL of 2.00 M H2SO4 required to neutralize 50.0 mL of 1.00 M KOH. H2SO4 + 2KOH K2SO4 + 2H2O 1) 12.5 mL 2) 50.0 mL 3) 200. mL

15. Solution Calculate the mL of 2.00 M H2SO4 required to neutralize 50.0 mL of 1.00 M KOH. H2SO4 + 2KOH K2SO4 + 2H2O 1) 12.5 mL 0.0500 L x 1.00 mole KOH x 1 mole H2SO4 x 1 L 2 mole KOH 1 L x 1000 mL = 12.5 mL 2 mole H2SO4 1 L

16. Learning Check A 25 mL sample of phosphoric acid is neutralized by 40. mL of 1.5 M NaOH. What is the molarity of the phosphoric acid solution? 3NaOH + H3PO4 Na3PO4 + 3H2O 1) 0.45 M 2) 0.80 M 3) 7.2 M

17. Solution 2) 0.80 M 0.040 L x 1.5 mole NaOH x 1 mole H3PO4 1 L 3 mole NaOH = 0.020 mole H3PO4 0.020 mole H3PO4 = 0.80 mole/L = 0.80 M 0.025 L