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V. Acid-Base Titration Titration is the process of adding a measured volume of an acid or base of known molarity (the standard solution) to an acid or.

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Presentation on theme: "V. Acid-Base Titration Titration is the process of adding a measured volume of an acid or base of known molarity (the standard solution) to an acid or."— Presentation transcript:

1 V. Acid-Base Titration Titration is the process of adding a measured volume of an acid or base of known molarity (the standard solution) to an acid or base of unknown molarity until neutralization occurs. standard solution – known molarity (concentration) unknown molarity An indicator is added to the flask. It changes color when neutralization has occurred.

2 Remember: acid + base  water +salt The net ionic equation of what really reacts is: H + + OH -  H 2 O Neutralization occurs when the number of moles of = number of moles of H + from the acid OH - from the base

3 molarity = moles solute liters of soln. molarity = moles volume molarity x vol. = moles

4 Remember: acid + base  water +salt The net ionic equation of what really reacts is: H + + OH -  H 2 O Neutralization occurs when the number of moles of = number of moles of H + from the acid OH - from the base (molarity x volume) acid = (molarity x volume) base M A V A = M B V B (in Table T)

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6 Remember: acid + base  water +salt The net ionic equation of what really reacts is: H + + OH -  H 2 O Neutralization occurs when the number of moles of = number of moles of H + from the acid OH - from the base (molarity x volume) acid = (molarity x volume) base M A V A = M B V B (in Table T) Volumes V A and V B must be in the same units (mL or L).

7 Ex. How many milliliters of 4.00 M NaOH are required to neutralize 50.0 milliliters of 2.00 M solution of HNO 3 ? M A V A = M B V B (2.00 M)(50.0 mL) = (4.00 M) x (2.00 M)(50.0 mL) = x 4.00 M 25 mL = x Note: 2x the concentration  ½ as much needed

8 Ex. What is the molarity of an HCl solution if 30. mL of the acid are needed to neutralize 10. mL of a 6.0 M NaOH solution? M A V A = M B V B x (30. mL) = (6.0 M)(10. mL) x = (6.0 M)(10. mL) 30. mL x = 2.0 M Note: 3x as much needed  1/3 the concentration

9 In the equation M A V A = M B V B, M A means moles of H + /liter 1 M HCl yields (gives) 1 mole of H + /liter 2 M HCl yields 2 moles of H + /liter Note: Diprotic acids yield twice as many H + /liter 1 M H 2 SO 4 yields 2 moles of H + /liter 2 M H 2 SO 4 yields 4 moles of H + /liter And:M B means moles of OH - /liter 1 M NaOH yields 1 mole of OH - /liter 2 M NaOH yields 2 moles of OH - /liter Note: Dihydroxy bases yield twice as many OH - /liter 1 M Ca(OH) 2 yields 2 moles of OH - /liter 2 M Ca(OH) 2 yields 4 moles of OH - /liter

10 Ex. What is the concentration of a sulfuric acid solution if 50. mL of 0.40 M KOH are needed to neutralize 20. mL of the H 2 SO 4 solution of unknown concentration? M A V A = M B V B x (20. mL) = (0.40 M)(50. mL) x = (0.40 M)(50. mL) 20. mL x = 1.0 M But H 2 SO 4 is diprotic. Each molecule produces 2 H+ ions. So the concentration only needs to be half as much: true x = 1.0 M/2 = 0.50 M

11 Experi- mental design: 2.A volume of known molarity is added to a burette and its initial volume is recorded. 1.A known volume of unknown molarity is measured and added to the flask. Then an indicator is added. 3.The burette solution is titrated into the flask solution until the indicator changes color 4.The volume of burette solution used is found by subtracting initial and final burette volumes

12 Ex. 75 mL of KOH of unknown concentration are added to a flask and some indicator is added. The burette contains 2.5 M HCl. Its initial volume reading is 19 mL. The acid is slowly added until the indicator changes color. The final burette volume is 61 mL. Find the concentration (molarity) of the KOH. M A V A = M B V B (2.5 M)( mL) = x (75 mL) (2.5 M)(42 mL) = x 75 mL 1.4 M = x


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