Vocab the basics first Strong Acids and Bases Dissociate 100 in water Acids CBS PIN HCl HBr H 2 SO 4 HClO 4 HI HNO 3 Bases Group IA LiCs and IIA MgBa hydroxides ID: 694566
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Slide1
Solution stoichiometry
Problems and solutionsSlide2
Vocab: the basics first
Strong Acids and Bases: Dissociate 100% in water
Acids: CBS PIN HCl HBr H2SO4 HClO4 HI HNO3 Bases: Group IA (Li-Cs) and IIA (Mg-Ba) hydroxidesMg(OH)2 Mg2+ + 2OH- H2SO4 H+ + HSO4- Slide3
Example #1
Dilution
If you need to make 75ml of 1.5M
HCl solution from 12M stock HCl. How much stock solution and water are needed? M1V1 = M2V2 75ml(1.5M) = (12M)(V2) 9.4ml of stock and 65.6ml H2OSlide4
Example #2
Stoichiometry
What volume of Na
3PO4 (6.0M) is needed to react with 120ml of 4.2M ZnCl2?Both reactants are soluble: Na3PO4 3Na+ + PO43- ZnCl2 2Cl- + Zn2+ Net ionic: 3Zn2+ + 2PO43- Zn3
(PO4)2Slide5
Example #2…cont.
Net ionic:
3Zn2+ + 2PO43- Zn3(PO4)2Use stoichiometry:.120L 4.2molZnCl2 1molZn2+ 2molPO43- 1molNa3PO4 1L = 1L 1molZnCl
2 3molZn2+ 1molPO43-
6molNa
3
PO
4
=
.056L = 56ml Na
3PO
4
Slide6
Example #3
A 1.42g sample of a pure compound, with the formula M
2
SO4, was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected and found to have a mass of 1.36g. Determine the atomic mass of M and identify MSlide7
Example #3 –
Just one method of solving
M
2SO4 + CaCl2 2MCl + CaSO4 1.42g 1.36g1.36gCaSO4 1molCaSO4 1molM2SO4 Xg = 136g 1molCaSO4 1mol M2SO4.01x g = 1.42g M2SO
4X=142g M
2
SO
4
subtract the mass of SO
4
2M = 45.94g therefore, mass M = 22.97g which is NaSlide8
Example #4
Consider a 1.50g mixture of magnesium nitrate and magnesium chloride. After dissolving this mixture in water, 0.500M silver nitrate is added drop-wise until precipitate formation is complete. The mass of
the white
precipitate formed is 0.641g.Calculate the mass percent of magnesium chloride in the mixtureDetermine the minimum volume of silver nitrate that must be added to ensure the complete formation of the precipitate.Slide9
Example
#4
–
Just one method of solvingMgCl2 + 2AgNO3 2AgCl + Mg(NO3)21.50g mixture 0.500M .641g.641gAgCl 1molAgCl 1molMgCl2 95.21g = .213gMgCl2 143.15g 2mol AgCl 1molMgCl2 .213gMgCl2 / 1.50g mixture = 14.2%.213gMgCl2 1molMgCl2 2molAgNO3
1L = 8.95ml AgNO3 95.21g 1molMgCl
2
.500M
AgNO
3Slide10
Example #5
A 230. ml sample of 0.275M CaCl
2
solution is left on a hotplate overnight. The following morning, the solution is 1.10M.What volume of water evaporated from the 0.275M CaCl2 solution?Slide11
Example
#5
–
Just one method of solvingCaCl2 .275M 230ml (.230L) .275M = Xmol / .230L .0633mol CaCl2 1.10M = .0633mol / XL .0575L = vol in morning230ml-57.5ml = 173ml evaporatedSlide12
Example #6
Mixing
Given 120ml of 2.2M
LiCl and 400ml of 4.2M AlCl3, what is the concentration of Cl- after mixing?Write dissociation reactions: LiCl Li+ + Cl- AlCl3 Al3+ + 3Cl- Determine the moles of Cl
- in each reaction:.120L 2.2mol LiCl 1mol
Cl
-
= 0.264
mol
Cl
- 1L 1mol LiClSlide13
Example #6….cont.
.400L 4.2mol AlCl
3
3mol Cl- = 5.04mol Cl- 1L 1mol AlCl3 Add moles of Cl- together = 5.304 mol Cl-Divide moles by the total volume of the solution. 5.304mol Cl- / .520L = 10.2M