 # Warmup (10 minutes) 1. What is the molarity of a solution where 39.9 g CuSO 4 are dissolved in 250.ml water? 39.9g(1mole)/159.62g = 0.24997 moles 0.24997.

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Warmup (10 minutes) 1. What is the molarity of a solution where 39.9 g CuSO 4 are dissolved in 250.ml water? 39.9g(1mole)/159.62g = 0.24997 moles 0.24997 moles/0.250L = 1.00M 2. Check out these flasks: a. Compare the amount of solute in each flask. b. If only one solution was prepared from the salt to make the rest, which was it?

Dilutions and Titrations and Solution Stoichiometry

Solution Preparation from a Solid 1. How would you prepare a 1.00 M solution of CuSO 4 in a 250. mL volumetric flask? 1. Figure out mass of solid you need to dissolve in that particular amount of water. M = mole L 1.00M = mole 0.250 L 0.250 moles CuSO 4 1 mole CuSO 4 (159.62 g CuSO 4 ) = 39.9 g CuSO 4 2. Weigh out that mass of solid, place in flask. 3. Add about half the amount of water, swirl to dissolve solute. 4. Add rest of water – do not exceed the agreed- upon solution volume!

Solution Preparation by Diluting a Concentrated (Stock) Solution 2. How would you prepare 250.ml of a 0.300 M CuSO 4 solution using the 1.00M solution we just made? 1. Figure out what volume of the stock solution that you need draw out to add to more water for the dilution. Formula:M 1 V 1 = M 2 V 2 V 1 = M 2 V 2 M 1 V 1 = (0.300 M)(250.ml) 1.00M V 1 = 75.0 ml 2. Add 75.0 ml of the 1.00M stock solution to 175 ml water (75.0ml + 175ml = 250.ml)

3. If 0.15 L of a 6.00 M H 2 SO 4 solution is diluted to make a 0.50 L solution, what is the molarity of the diluted H 2 SO 4 solution? M 1 V 1 = M 2 V 2 M 1 V 1 = M 2 V 2 M 2 = (6.00 M)(0.15L) 0.50 L M 2 = 1.8 M If the diluted solution is calculated to be MORE CONCENTRATED than the stock solution…. there is a problem!!!!

Titrations 4. Suppose that 15.0 mL of a 2.50 x 10 -2 M solution of HCl is required to neutralize 10.0 mL of a KOH solution. What is the molarity of the KOH solution? Neutralization means the resulting pH = 7 M a V a = M b V b (2.50 x 10 -2 M)(15.0ml) = M b (10.0ml) 3.75 x 10 -2 M = M b

5. A titration is performed using by adding 0.100 M NaOH to 40.0 mL of 0.300 M of HCl (hydrochloric acid). Calculate the volume of base needed to reach the equivalence point. moles H+ = moles OH- neutralization! Oversimplification: moles acid = moles base

5. A titration is performed using by adding 0.100 M NaOH to 40.0 mL of 0.300 M of a strong acid. Calculate the volume of base needed to reach the equivalence point. Translation: “how much volume for neutralization” or “how much base needs to be added so that the moles base added to the flask equals to the moles acid in the flask?” M a V a = M b V b (0.300M)(40.0ml) = (0.100M)( V b ) 120. ml = V b

6. A concentration less than 0.0050M Cu + in our drinking water is considered safe. A titration is performed on a 25.0 ml water sample to determine the concentration of copper(I) ions. After 13.2 ml of a 0.250M NaI solution is added, a precipitate forms. Upon further addition of NaI, no more precipitate forms. What is the [Cu + ] in the sample? Na + + I - + Cu +  ???????? I - (aq) + Cu + (aq)  CuI(s) M i V i = M Cu V Cu 0.132M = M Cu (0.250M)(13.2ml) = M Cu (25.0ml)

Solution Stoichiometry 7. 60.0 mL of a 0.050 M solution of AgNO 3 is mixed with 40.0 mL of a 0.080 M solution of KCl. a.Write the equation showing the precipitate formation. KCl(aq) + AgNO 3 (aq)  b. How many moles of each ion are present? c. How many moles of precipitate will form? K + + Cl - + Ag + + NO 3 - Cl - (aq) + Ag + (aq)  AgCl(s) 0.0032 0.0030 moles Because Ag + is limiting, only 0.0030 moles ppt can form. 0.0002 moles of Cl - will be leftover. 0.050M = x moles AgNO 3 (or Ag+) 0.0600 L 0.080 M = x moles KCl (or Cl-) 0.040 L

8. 5.67 grams calcium nitrate is used to make a 2.00M Ca(NO 3 ) 2 solution. This solution is mixed with 20.0 ml of a 1.0M Na 2 CrO 4 solution. a. Calculate the volume of the Ca(NO 3 ) 2 solution 2.00M = mole x L 5.67 g Ca(NO 3 ) 2 1 mole Ca(NO 3 ) 2 164.10 g Ca(NO 3 ) 2 = 0.0346 moles Ca(NO 3 ) 2 2.00M = 0.0346 moles x L 2.00M (x)= 0.0346 moles x = 0.0173 L or 17.3 ml b. How many moles of sodium chromate are in the solution? 1.00M = mole 0.0200L 0.0200 moles Na 2 CrO 4

8. 5.67 grams calcium nitrate is used to make a 2.00M Ca(NO 3 ) 2 solution. This solution is mixed with 20.0 ml of a 1.0M Na 2 CrO 4 solution. c. How many moles of precipitate can form when the two solutions are combined? Na 2 CrO 4 (aq) + Ca(NO 3 ) 2 (aq)  d. What mass of precipitate will be left when all the water has evaporated? Na + + CrO 4 2- + Ca 2+ + NO 3 - CrO 4 2- (aq) + Ca 2+ (aq)  CaCrO 4 (s) 0.0200 0.0346 moles 0.0200 moles CaCrO 4 1 mole CaCrO 4 156.08 g CaCrO 4 = 3.12 g CaCrO 4 only 0.0200 moles ppt can form (chromate is limiting)

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