1. The Battle for Chemical Domination On the Open Seas
Group Slaughter
The Battle for Chemical Domination
On the Open Seas

2. Set-Up: Number 1-10 on your piece of paper.
Each team member must print their name next to ONE number only.
Don’t let any other team see what your numbers are! These are your battleships.
If your team has less than 4 members, add sufficient names to reach 4 total people.

3. The Rules
You gain the ability to take shots at other teams’ battleships by getting questions correct. Questions are worth 1, 2, or 3 shots at the team of your choice.
When one of your battleships is sunk, the person who’s battleship was shot will be rescued by their teammates. THE TEAM ONLY DIES WHEN EVERY MEMBER HAS BEEN SHOT.
Once a team is dead, if they get the next question right, they will come back to life as angry ghosts! Ghosts get double shots for every question they get right.

4. The Rules: Part II
Winners will be determined by which team has the most LIVING battleships at the end of the class.
All students MUST participate. You show this by making sure everyone’s color appears in the answer.
Sharing pens results in immediate disqualification for that question.
Answers are only accepted if written on the group’s white board!
NO FRIENDLY FIRE ;)

5. May the odds be ever in your favor…
Last Questions?
Let the games begin!
May the odds be ever in your favor…

6. Question 1: for 1 shot
Write a balanced equation for the neutralization reaction between hydrophosphoric acid and magnesium hydroxide.

7. Question 1: for 1 shot 2 H3P + 3 Mg(OH)2  6 H2O + Mg3P2
1 Write a balanced equation for the neutralization reaction between hydrophosphoric acid and calcium hydroxide.
2 H3P + 3 Mg(OH)2  6 H2O + Mg3P2

8. Question 2: for 1 shot
25.0 mL of 0.12 M HBr were used to completely neutralize 15.0 mL of RbOH in a titration.
What was the initial concentration of RbOH?

9. Question 2: for 1 shot
25.0 mL of 0.12 M HBr were used to completely neutralize 15.0 mL of RbOH in a titration.What was the initial concentration of RbOH?
MaVa=MbVb
Mb = 25.0 x 0.12 / 15.0 = 0.20 M RbOH

10. Question 3: for 2 shots What was the initial pH of the RbOH solution?
25.0 mL of 0.12 M HBr completely neutralized 15.0 mL of RbOH in a titration, and the initial [RbOH] was determined to be 0.20 M.
What was the initial pH of the RbOH solution?
What was the initial pH of the HBr solution?

11. Question 3: for 2 shots
25.0 mL of 0.12 M HBr completely neutralized 15.0 mL of RbOH in a titration, and the initial [RbOH] was determined to be 0.20 M.
What was the initial pH of the RbOH solution?
pOH = - log(0.20) = 0.70, pH = 14 – 0.70 = 13.30
What was the initial pH of the HBr solution?
pH = - log(0.12) = 0.92

12. Question 4: for 2 shots
25.0 mL of 0.12 M HBr (pH = 0.92) completely neutralized 15.0 mL of RbOH in a titration. The initial [RbOH] (the analyte) was determined to be 0.20 M (pH = 13.3).
Sketch a graph of the titration curve which would result from this experiment, including:
Labeled axes
Labeled equivalence point

13. Question 4: for 2 shots
25.0 mL of 0.12 M HBr (pH = 0.92) completely neutralized 15.0 mL of RbOH (pH = 13.30) in a titration, and the initial [RbOH] was determined to be 0.20 M. Sketch a graph of the titration curve, including:
Labeled axes
Labeled equivalence point

14. Question 5: for 1 shot
Which acid shown below at equilibrium is stronger, and why?
Acid Acid 2

15. Question 5: for 1 shot
Which acid shown below is stronger, and why?
Acid Acid 2
Acid 1 is stronger, because more of the acid particles (HA) dissociate into ions (H+ and A-)

16. Question 6: for 1 shot Identify AND label from the rxn below:
 both conjugate acid/base pairs!
HC2H3O2 + HCO3− ⇋ C2H3O2− + H2CO3

17. Question 6: for 1 shot HC2H3O2 + HCO3− ⇋ C2H3O2− + H2CO3
Identify AND label both conjugate acid/base pairs in the reaction below:
HC2H3O2 + HCO3− ⇋ C2H3O2− + H2CO3
Acid Base CB CA
HC2H3O2 /C2H3O2− and HCO3− / H2CO3
Acid / conjugate base Base / conjugate acid

18. Question 7: for 2 shots HN3 + ClO2− ⇋ N3− + HClO2
Answer based on the Ka values provided.
Is this reaction reactant-favored or product-favored?
Will K be less than, equal to, or greater than 1?
Explain how you answered parts (a) and (b).
HN3 + ClO2− ⇋ N3− + HClO2
Ka of HN3 = 2.5 x 10−5
Ka of HClO2 = 1.1 x 10−2

19. Question 7: for 2 shots Ka of HN3 = 2.5 x 10−5
reactant-favored
K < 1
HN3 is the weaker acid (smaller Ka), and the favored reaction direction is when the weaker acid/base is produced.
HN3 + ClO2− ⇋ N3− + HClO2
Ka of HN3 = 2.5 x 10−5
Ka of HClO2 = 1.1 x 10−2

20. Question 8: for 1 shot
At 25oC, Kw is a constant which equals 1.0 x 10−14. What variable would you need to adjust to change the value of Kw?
The concentration of H+.
The concentration of OH−.
The temperature.
Both (a) and (b).
All of the above: (a), (b), and (c).

21. Question 8: for 1 shot The temperature.
At 25oC, Kw is a constant which equals 1.0 x 10−14. What variable would you need to adjust to change the value of Kw?
The concentration of H+.
The concentration of OH−.
The temperature.
Both (a) and (b).
All of the above: (a), (b), and (c).

22. Question 9: for 2 shots
An Al(OH)3 solution has a pH of Determine the concentration of Al(OH)3.

23. Question 9: for 2 shots
An Al(OH)3 solution has a pH of Determine the concentration of Al(OH)3.
pOH = 14 – 4.94 = 9.06
[OH −] = 10−9.06 = 8.7 x 10−10
[Al(OH)3] = 8.7 x 10−10 / 3 = 2.9 x 10−10 M

24. Question 10: for 1 shot Complete the chart below: Acids Bases
Arrhenius
Brønsted-–Lowry
Acids
____ donor
Bases
OH− __________
H+ __________

25. Question 10: for 1 shot Complete the chart below: Acids H+ donor Bases
Arrhenius
Brønsted-–Lowry
Acids
H+ donor
Bases
OH− donor
H+ acceptor

26. 3 NaOH(aq) + H3PO4(aq) → 3 H2O(l) + Na3PO4(aq)
Question 11: for 1 shot
What is the net ionic equation for the neutralization reaction shown below?
3 NaOH(aq) + H3PO4(aq) → 3 H2O(l) + Na3PO4(aq)

27. H+(aq) + OH− (aq) → H2O(l)
Question 11: for 1 shot
What is the net ionic equation for the neutralization reaction shown below?
3 NaOH(aq) + H3PO4(aq) → 3 H2O(l) + Na3PO4(aq)
H+(aq) + OH− (aq) → H2O(l)

28. Question 12: for 1 shot
By titration, what volume of 0.65 M H2SO4 can be neutralized by 38 mL of 0.40 M LiOH solution?

29. Question 12: for 1 shot MH+Va=MbVb Va = 0.40 x 38 / (0.65 x 2) = 12 mL
By titration, what volume of 0.65 M H2SO4 can be neutralized by 38 mL of 0.40 M LiOH solution?
MH+Va=MbVb
Va = 0.40 x 38 / (0.65 x 2) = 12 mL