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- 1. Ch 21.1 Neutralization Reactions
- 2. Acid-Base Reactions • Acid + Base Water + Salt • Properties related to every day: – antacids depend on neutralization – farmers use it to control soil pH – formation of cave stalactites – human body kidney stones from insoluble salts
- 3. Acid-Base Reactions • Neutralization Reaction - a reaction in which an acid and a base react in an aqueous solution to produce a salt and water: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2 H2O(l) – Table 21.1, page 614 lists some salts
- 4. Titration • Titration is the process of adding a known amount of solution of known concentration to determine the concentration of another solution • Remember? - a balanced equation has a mole ratio
- 5. Titrations BuretBuret - volumetric glassware used for titrations. It allows you to add a known amount of your titrant to the solution you are testing. If a pH meter is used, the equivalence point can be measured. An indicator will give you the endpoint.
- 6. Titrations • Note the color change which indicates that the ‘endpoint’ has been reached. Start End
- 7. - Page 614
- 8. Titration • The concentration of acid (or base) in solution can be determined by performing a neutralization reaction –An indicator is used to show when neutralization has occurred –Often we use phenolphthalein- because it is colorless in neutral and acid; turns pink in base
- 9. Steps - Neutralization reaction #1. A measured volume of acid of unknown concentration is added to a flask #2. Several drops of indicator added #3. A base of known concentration is slowly added, until the indicator changes color; measure the volume – Figure 21.4, page 617 – ASAP Sim26
- 10. Neutralization • The solution of known concentration is called the standard solution – added by using a buret • Continue adding until the indicator changes color – called the “end point” of the titration – Sample Problem 21.2, page 618
- 11. Equivalents • An equivalent is the amount of acid or base that will give 1 mole of hydrogen ions, or in the case of a base, hydroxide ions ( remember not all acids and bases are the same strength and so, a 1M solution of an acid may or may not give 1 mole of hydrogen ions) Example; 1 mole of HCl is 1 equivalent and will neutralize 1 mole of NaOH, which is also 1 equivalent BUT, 1 mole of H2SO4 has 2 equivalents and will neutralize 2 moles of NaOH
- 12. Equivalents • The mass of one equivalent of a substance is called the gram equivalent mass • The gram equivalent mass for HCl is 36.5 g/mol • The gram equivalent mass for H2SO4 is 49g/mol (1 mole H2SO4 is 2 equivalents, so 1 equivalent is ½ mole)
- 13. Equivalent Problem • What is the mass of 1 equivalent of Ca(OH)2? Molar mass is 74g/mol 1mole Ca(OH)2 has 2 equivalents 74g/mol = 37g/eq 2 eq./mol
- 14. Equivalent problem • How many equivalents are in 14.6 g of sulfuric acid (H2SO4) 1mole H2SO4 has 2 equivalents Molar mass is 98.1g/mol So… 98.1g/2eq = 49.1g/eq 14.6 g x 1eq/49.1g=.297 eq
- 15. Normality • Normality (N)=equivalent/L • IF 1 mole of acid or base gives 1 equiv. then Molarity (M) will equal Normality (N) • If 1 mole of acid =2 equiv,, then N=2M; if 1 mole of acid =3 equiv, then N=3M • If you know the normality you can find the # of equiv. by. Equiv= V (in L) x N
- 16. Normality problem • How many equiv. are in 2.5L of 0.60N H2SO4? Equiv=V x N Equiv=2.5L x 0.60N=1.5 equiv.
- 17. Normality & Titrations • If you are diluting a solution of known normality, or in the case of titration you will find the following formula useful: V1N1=V2N2 In the case of titrations VaNa=VbNb
- 18. Problem Using V1N1=V2N2 • You need to make 250mL of 0.10N sodium hydroxide from a stock solution of NaOH that is 2.0N. How many mL of the stock solution do you need? V1N1=V2N2 250mLx 0.10N=V2x2.0N 250mLx 0.10N = V2 2.0N 13mL=V2
- 19. • If a 35.0mL of a 0.20N HCl are needed to neutralize 25.0mL of an unknown base, what is the normality of the base? VaNa=VbNb 35.0mL x 0.20N =25.0mLxNb 35.0mL x 0.20N=Nb 25.0mL 0.28N=Nb Problem Using V1N1=V2N2
- 20. • How many mL of 0.500N sulfuric acid are needed to neutralize 50.0mL of 0.200N potassium hydroxide? VaNa=VbNb Va x 0.500N = 50.0mL x 0.200N Va= 50.0mL x 0.200N 0.500N Va=20.0mL Problem Using V1N1=V2N2

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