1 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: HNO 3 + LiOH  LiNO 3 + H 2 0

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 Requires a standard solution an acid-base indicator and a graduated burette Typically the indicator used is Phenolphthalein when titrating strong acids and bases Typically the indicator used is Phenolphthalein when titrating strong acids and bases A Standard solution is an acid or base of known molar concentration. A Standard solution is an acid or base of known molar concentration. If titrating an acid we need a standardized base. If titrating an acid we need a standardized base. If titrating a base we need a standardized acid 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. The Standard solution is slowly added to the unknown solution drop by drop. As the solutions mix, a neutralization reaction occurs. As the solutions mix, a neutralization reaction occurs. Eventually, enough standard solution is added to neutralize the unknown solution. This is the Equivalence point. Eventually, enough standard solution is added to neutralize the unknown solution. This is the Equivalence point. Example: 1 mol H 2 SO mol KOH  1 mol K 2 SO mol H 2 0

7 1 ml Reading A Buret Initial reading 2.4 ml Final Reading4.8 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 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 - 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? mol ea.

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

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

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 M 1 V 1 = M 2 V 2 For example if M 1 = 2M and V 1 =100ml and it is then Diluted with 100ml of H 2 0 what is the new Molarity? 2M x 100ml = M 2 x 200ml M 2 = 200/200 = 1M

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

13 and M a V a = M b V b True for strong monoprotic acids and monohydroxy bases Since both H + and OH - (S#) =1. True for strong monoprotic acids and monohydroxy bases Since both H + and OH - (S#) =1. Or strong diprotic and dihydroxy acids & bases since both H + and OH - (S#) = 2 Or strong diprotic and dihydroxy acids & bases since both H + and OH - (S#) = 2 Like HCl & KOH Or H 2 SO 4 Ca(OH) 2 Like HCl & KOH Or H 2 SO 4 & 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 M NaOH is added. In a titration of 40.0 mL of a nitric acid solution, the end point is reached when 35.0 mL of M NaOH is added. Calculate the concentration of the nitric acid solution. Calculate the concentration of the nitric acid solution.

15 Neutralization Reaction HNO 3 + NaOH  H 2 O + NaNO 3 HNO 3 + NaOH  H 2 O + NaNO 3 HNO 3 is a strong monoprotic acid. HNO 3 is a strong monoprotic acid. NaOH is a strong monohydroxy base NaOH is a strong monohydroxy base

16 Variables M a = ? M a = ? V a = 40.0 mL V a = 40.0 mL M b = M M b = M V b = 35.0 mL V b = 35.0 mL # = 1 and 1 # = 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 M KOH solution are needed to neutralize 20.0 mL of the H 2 SO 4 solution of unknown concentration? What is the concentration of a sulfuric acid solution if 50.0 mL of a M KOH solution are needed to neutralize 20.0 mL of the H 2 SO 4 solution of unknown concentration?

19 Neutralization Reaction KOH + H 2 SO 4  H 2 O + K 2 SO 4 KOH + H 2 SO 4  H 2 O + K 2 SO 4 H 2 SO 4 is a strong Diprotic acid. H 2 SO 4 is a strong Diprotic acid. KOH is a strong monohydroxy base KOH is a strong monohydroxy base # H+ = 2, # OH- = 1) # H+ = 2, # OH- = 1)

20 Variables M a = ? M a = ? V a = 20.0 mL V a = 20.0 mL M b = M M b = M V b = 50.0 mL V b = 50.0 mL S A =2 & S B =1 S A =2 & S B =1

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