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?
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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.
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Warmup (10 minutes)1. What is the molarity of a solution where 39.9 g
CuSO4 are dissolved in 250.ml water?
39.9g(1mole)/159.62g = 0.24997 moles0.24997 moles/0.250L = 1.00M2. 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 Solid1. How would you prepare a 1.00 M solution of CuSO4 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 CuSO4
1 mole CuSO4
(159.62 g CuSO4 ) = 39.9 g CuSO4
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 CuSO4 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: M1V1 = M2V2V1 = M2V2
M1V1 = (0.300 M)(250.ml)1.00M
V1 = 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 H2SO4 solution is diluted to make a 0.50 L solution, what is the molarity of the diluted H2SO4 solution?
M1V1 = M2V2M1V1 = M2
V2
M2 = (6.00 M)(0.15L)
0.50 L
M2 = 1.8 MIf the diluted solution
is calculated to be MORE CONCENTRATED than the stock solution…. there is a problem!!!!
Titrations4. 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
MaVa = MbVb
(2.50 x 10-2 M)(15.0ml) = Mb(10.0ml)
3.75 x 10-2 M = Mb
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?”
MaVa = MbVb
(0.300M)(40.0ml) = (0.100M)(Vb)120. ml = Vb
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) MiVi = MCuVCu
0.132M = MCu
(0.250M)(13.2ml) = MCu (25.0ml)
Solution Stoichiometry7. 60.0 mL of a 0.050 M solution of AgNO3 is mixed with 40.0 mL of a 0.080 M solution of KCl.
a.Write the equation showing the precipitate formation.KCl(aq) + AgNO3(aq)