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Energetics and Thermodynamics - Basic Chemical Equations

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1.
a. Calculate the [H+] of a weak acid solution with [OH-] = 3.50x10^-8M
b. Calculate the pH of the solution
c. Calculate the [H+] of a Classic Coke if the pH=3.75

2. Write the balanced equations: (Neutralizations) in ionic form and show net ionic equation.
a. H2SO4(aq) + NaHCO3(aq)-->
b. H3PO4(aq)+Mg(OH)2-->

3. Calculate the molarity of a phosphoric acid solution if 50.25mL neutralizes 60.50mL of A 0.35 molar potassium hydroxide solution in an acid/base titration lab.

4. If 0.581 grams of oxalic acid (H2C2O4*2H2O requeires 30.52mL of KOH(aq) in a titration, calculate the molarity of the KOH solution

5. Calculate the calories needed to heat 1500 grams of ice from 20C to steam at 100C

6. An intravenous saline solution is 0.90% NaCl. What mass of NaCl should be added to 1.25 kilograms of water for the solution?

7. What is the mass of HCL (36.5g/mole) in 200mL of 12 molar hydrochloric acid?

8. How many kilograms of antifreeze(Ethylene Glycol), C2H6O2, must be added to 25 kilograms of water in order to lower the freezing point(delta T) of the truck's radiator fluid to -25C (Kfp=-1.86C/molal)

9. Calculate the molar concentration (molarity) of a solution containing 200g of glucose (C6H12O6) in 300mL of solution.

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1.
a. if the [OH-]=3.5x10-8 and we know that [H+]x[OH-]=10^-14, the by division we get [H+]=2.88x10^-7
b. if the [OH-]=3.5x10-8 , then the pOH= -log[OH-]=7.46. we also know that pOH+pH=14 so pH=6.54
c. is pH= 3.75, then [H+]=10^-pH=1.77x10^-4

2.

a. H2SO4+2NaHCO3  Na2SO4+ 2H2CO3  2H2O + 2CO2
since Na and SO4 stay the same on both sides (ions) we can cancel on both sides leaving 2H+ + 2HCO3 2H2CO3  H2O+ CO2
since H2CO3 is a very weak acid under normal conditions we get an equilibrium between the acid form and water and CO2 as shown.

b. Mg and PO4-3 are both spectator ions... meaning as above they remain ions on both sides of the reaction without undergoing condensation, or anything else. And so the neutralization reaction becomes:
6H+ + 6OH-  3H2O (note that we have balanced both sides by multiplying both chemicals by the appropriate factor (2 and 3))

3.
Phosphoric acid is H3PO4 (it supplies 3 ...

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