1. Titration A lab technique to determine the molarity of a compound

2. Basic premise… A measured volume of a solution of unknown molarity is put into a flask Another chemical of known concentration is slowly dripped from a buret into the unknown and a reaction ensues The reaction proceeds until it is indicated the reaction has reached completion Using stoichiometry, the molarity of the unknown is ascertained

3. Basic setup

4. indicator Usually, visual indication that the reaction has reached completion is used A chemical indicator is one that changes color when the reaction has reached completion, known as the “end point” Example: litmus is red in acids and blue in bases colorlessPhenolphthalein is colorless in acids and pink in bases

5. titration lab calculation 1.Read the volume of the base dripped in from the buret -get to moles: M x L = moles of OH- 2.Use the coefficients of the balanced equation to convert the moles of OH- dripped in from the buret into the moles of acid with an unknown molarity in the flask -Switch the moles: moles of OH- x (#moles acid / #moles base) 3.Knowing the volume of the sample of acid that was titrated, convert the moles of acid into it’s molarity -get out of moles: [acid] = moles/L 4.Knowing that pH = -log[H + ] calculate the pH of the original acid solution

6. titration lab calculation Typical titration question: If it takes 17.4mL of 0.05M NaOH to titrate a 25mL sample of HCl to its endpoint, what is the molarity of the acid, and what is it’s pH? Write the balanced equation: NaOH + HCl  H 2 O + NaCl

7. titration lab calculation 1.Read the volume of the base dripped in from the buret -get to moles: 0.05M x 0.0174L = 8.7x10 -4 moles of base 2. Use the coefficients of the balanced equation to convert the moles of OH- dripped in from the buret into the moles of acid with an unknown molarity in the flask 8.7x10 -4 mol of OH- x (1 mol acid/ 1 mol base) = 8.7x10 -4 mol of acid 3. Knowing the volume of the sample of acid that was titrated, convert the moles of acid into it’s molarity -get out of moles: [acid] = 8.7x10 -4 mol / 0.025 L = 0.035M 4. Knowing that pH = -log[H + ] calculate the pH of the original acid solution: pH = -log[0.035] = 1.5

8. titration lab calculation Typical titration question: If it takes 9.4mL of 0.01M HBr to titrate a 20mL sample of Ca(OH) 2 to its endpoint, what is the molarity of the base, and what is it’s pH? Write the balanced equation: Ca(OH) 2 + 2 HBr  2H 2 O + CaBr 2

9. titration lab calculation 1.Read the volume of the acid dripped in from the buret -get to moles: 0.01M x 0.0094L = 9.4x10 -5 moles of acid 2. Use the coefficients of the balanced equation to convert the moles of acid dripped in from the buret into the moles of base with an unknown molarity in the flask 9.4x10 -5 mol of acid x (1 mol base/ 2 mol acid) = 4.7x10 -5 mol of base 3. Knowing the volume of the sample of acid that was titrated, convert the moles of acid into it’s molarity -get out of moles: [base] = 4.7x10 -5 mol / 0.020 L = 0.0024M

10. titration lab calculation 4. Because there are 2 OH - in each Ca(OH) 2, [OH - ] = 2x[Ca(OH) 2 ] 0.0024M x 2 = 0.0048M OH - 5. Knowing that pOH = -log[OH-] calculate the pOH of the original base solution: pOH = -log[0.0048] = 2.3 pH = 14 – 2.3 = 11.7