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Acids and Bases - Titration

Acids and Bases - Titration. And Acid/Base dilution. Mr. Shields Regents Chemistry U15 L05. Neutralization. Remember we said previously that one of the Property’s of acids and bases is that they react together Do you remember what this reaction is called?

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Acids and Bases - Titration

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  1. Acids and Bases - Titration And Acid/Base dilution Mr. Shields Regents Chemistry U15 L05

  2. Neutralization Remember we said previously that one of the Property’s of acids and bases is that they react together Do you remember what this reaction is called? It’s neutralization. So what exactly is neutralization? Neutralization occurs when acids and bases react. This results in 2 products – a salt and water. For example: HNO3 + LiOH  LiNO3 + H20

  3. Titration Titrations are carefully “Controlled” Neutralizations. A titration is used to determine the amount of acid or Base present in an unknown. - for example: you might want to know how acidic is in a sample of collected lake water.

  4. Acid-Base Titration • Requires a standard solution an acid-base indicator and a graduated burette • Typically the indicator used is Phenolphthalein when titrating strong acids and bases • A Standard solution is an acid or base of known molar concentration. • If titrating an acid we need a standardized base. • If titrating a base we need a standardized acid

  5. Recall Molarity (M) = # moles/liter of solution For example: If we add 0.5 moles of KOH to 750 ml of water what is the Molarity? M = 0.5 mol / 0.75L = 0.67M

  6. Titration • The Standard solution is slowly added to the unknown solution drop by drop. • As the solutions mix, a neutralization reaction occurs. • Eventually, enough standard solution is added to neutralize the unknown solution. This is the Equivalence point. Example: 1 mol H2SO4 + 2 mol KOH 1 mol K2SO4 + 2 mol H20

  7. 1 ml 1 ml 2 2 3 3 4 4 5 5 6 6 Final Reading 4.8 ml Reading A Buret Initial reading 2.4 ml Volume used = 2.4 ml

  8. Equivalence point • Total number of moles of H+ ions donated by acid = total number of moles of OH- donated by the base. • Total moles H+ = total moles OH- Ex: 50ml 0.5M HCL + 50ml 0.5M KOH have equal Amounts of H+ and OH- How many moles of HCl and KOH are there? 0.025 mol ea.

  9. Titration • End-point = point at which indicator changes color. • In titrating an unk. acid phenolphthalein changes from colorless to pink (what would the color change be if we were titrating a base with an acid?) • If the indicator is chosen correctly, the end-point is very close to the equivalence point.

  10. 14- Phenolphthalein Color change: 8.2 End point pH Between pH of 4 and 10, only a few drops of base are added. 7- Equivalence Pt  0-    Volume of 0.100 M NaOH added (ml) 0 ml 40ml 20 ml Titration of a strong acid with a strong base

  11. Calculation of Concentration Think back to our discussion of solutions. Do you Recall the equation for determining Molarity by Dilution? Calculating Molarity after dilution is accomplished very simply by using the equation M1V1 = M2V2 For example if M1 = 2M and V1=100ml and it is then Diluted with 100ml of H20 what is the new Molarity? 2M x 100ml = M2 x 200ml M2= 200/200 = 1M

  12. In neutralization we use a very similar equation #H+ (MH+VH+ )= #OH- (MOH-VOH-) MH+ = molarity of H+ # = Subscript in the MOH- = molarity of OH- acid or base for VH+ = volume of H+ H+ or OH- VOH- = volume of OH- Neutralization Equation

  13. and MaVa = MbVb • True for strong monoprotic acids andmonohydroxy bases Since both H+ and OH- (S#) =1. • Or strong diprotic and dihydroxy acids & bases since both H+ and OH- (S#) = 2 • Like HCl & KOH Or H2SO4 & Ca(OH)2

  14. Titration Problem #1 • In a titration of 40.0 mL of a nitric acid solution, the end point is reached when 35.0 mL of 0.100 M NaOH is added. • Calculate the concentration of the nitric acid solution.

  15. Neutralization Reaction • HNO3 + NaOH  H2O + NaNO3 • HNO3 is a strong monoprotic acid. • NaOH is a strong monohydroxy base

  16. Variables • Ma = ? • Va = 40.0 mL • Mb = 0.100 M • Vb = 35.0 mL • # = 1 and 1

  17. Plug and Chug • MA (40.0 mL) = (0.100 M )(35.0 mL) • X = Molarity (M) = .0875 M

  18. Titration Problem #2 • What is the concentration of a sulfuric acid solution if 50.0 mL of a 0.250 M KOH solution are needed to neutralize 20.0 mL of the H2SO4 solution of unknown concentration?

  19. Neutralization Reaction • KOH + H2SO4 H2O + K2SO4 • H2SO4 is a strong Diprotic acid. • KOH is a strong monohydroxy base • #H+ = 2, #OH- = 1)

  20. Variables • Ma = ? • Va = 20.0 mL • Mb = 0.250 M • Vb = 50.0 mL • SA =2 & SB =1

  21. Plug and Chug • #AxMAxVA = #BxMBxVB • 2xMA (20.0 mL) = 1x(0.250 M) (50.0 mL) • MA= 0.312 M

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