# 1 9.10 & 9.11 Titration Calculations Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings Chapter 9 Acids, Bases, & Salts Base (NaOH)

## Presentation on theme: "1 9.10 & 9.11 Titration Calculations Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings Chapter 9 Acids, Bases, & Salts Base (NaOH)"— Presentation transcript:

1 9.10 & 9.11 Titration Calculations Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings Chapter 9 Acids, Bases, & Salts Base (NaOH) Acid solution

2 Acid-Base Titration Titration is a laboratory procedure used to determine the molarity of an acid. uses a base such as NaOH to neutralize a measured volume of an acid. Base (NaOH) Acid solution Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

3 Indicator An indicator is added to the acid in the flask. causes the solution to change color when the acid is neutralized. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

4 End Point of Titration At the end point, the indicator gives the solution a permanent pink 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. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

5 Calculating Molarity What is the molarity of an HCl solution if 18.5 mL of a 0.225 M NaOH are required to neutralize 10.0 mL HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) STEP 1 Given: 18.5 mL of 0.225 M NaOH; 10.0 mL HCl Need: Molarity of HCl STEP 2 18.5 mL L moles NaOH moles HCl M HCl L HCl STEP 3 1 L = 1000 mL 0.225 mole NaOH/1 L NaOH 1 mole HCl/1 mole NaOH

6 Calculating Molarity (continued) STEP 4 Calculate the molarity of HCl. 18.5 mL NaOH x 1 L NaOH x 0.225 mole NaOH 1000 mL NaOH 1 L NaOH L moles NaOH x 1 mole HCl = 0.00416 mole HCl 1 mole NaOH M HCl = 0.00416 mole HCl = 0.416 M HCl 0.0100 L HCl

7 Calculate the mL of 2.00 M H 2 SO 4 required to neutralize 50.0 mL of 1.00 M KOH. H 2 SO 4 (aq) + 2KOH(aq) K 2 SO 4 (aq) + 2H 2 O(l) 1) 12.5 mL 2) 50.0 mL 3) 200. mL Learning Check

8 Solution 1)12.5 mL 0.0500 L KOH x 1.00 mole KOH x 1 mole H 2 SO 4 x 1 L KOH 2 mole KOH 1 L H 2 SO 4 x 1000 mL = 12.5 mL 2.00 mole H 2 SO 4 1 L H 2 SO 4

9 A 25.0 mL sample of phosphoric acid is neutralized by 42.6 mL of 1.45 M NaOH. What is the molarity of the phosphoric acid solution? 3NaOH(aq) + H 3 PO 4 (aq) Na 3 PO 4 (aq) + 3H 2 O(l) 1) 0.620 M 2) 0.824 M 3) 0.185 M Learning Check

10 Solution 2) 0.824 M 0.0426 L x 1.45 mole NaOH x 1 mole H 3 PO 4 1 L 3 mole NaOH = 0.0206 mole H 3 PO 4 0.0206 mole H 3 PO 4 = 0.824 mole/L = 0.824 M 0.0250 L

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