1....A solution is made by adding 580 grams of Potassium phosphate to 5.5 kilogr

Question

1....A solution is made by adding 580 grams of Potassium phosphate to 5.5 kilograms of water giving a final volme of 5.6 L.... wht is the a) mass percent b) molarity c) molality d) boiling point e) freezing point.....

2.....450 grams of nitrogen gas ha a pressure of 720 mm Hg and temp of 44C. what is the nitrogen gas VOlume?

3... H3PO4 + Ba(OH)2 ->???? a) complete equation b) whats the limiting reactant c)how many grams of barium containing prduct would be produced.

4...complete .... [H3O+] [OH-1] PH POH

3.88 X 10^-12_______________________

11.54

_________________________________

Explanation / Answer

1) Mass % = mass of solute / mass of solution

Mass % = .580 kg / 5.5 kg

Mass % = 0.10545 or 10.5%

Molarity = moles / vol L

Molarity = (580 g / 212.27 g) / 5.6 L

Molarity = 0.4879 or 0.49

Molality = moles solute / kg solvent

Molality = (580 g / 212.27 g) / 5.5 kg

Molality = 0.49679 or 0.50

Boiling Point

delta T = iKbm where i is the van't Hoff factor, Kb of water is 0.512 degrees Celcius, and m is molality

delta T = (4) (0.512)(0.50)

delta T = 1.024 degrees Celsius

Freezing Point

delta T = iKfm where i is the van't Hoff factor, Kf of water is 1.86 degrees Celcius, and m is molality

delta T = (4)(1.86)(0.50)

delta T = 3.72 degrees Celsius

2) V= nRT / p

= (450g / 28g)(0.08201)(317K) / (720 mmHg / 760 mmHg)

=441 atm

3) 3Ba(OH)2 (aq) + 2H3PO4 (aq) ---> Ba3(PO4)2 (aq) +6H2O (l)

18 grams of Phosphoric acid = 0.18 mol (divided by the formula weight of phosphoric acid, which is 98 g)

100 mL of 0.015 M Barium = 0.0015 mol (Use molarity formula to solve for moles, but remember to change 100 mL to .1 L)

We can now compare the molar ratios of these two reactants. Our formula shows us that we need 3 moles of Barium Hydroxide for every 2 moles of Phosphoric acid. If we start out with 0.18 mol of phosphoric acid, we would need 0.27 mol of Barium Hydroxide (I did this by taking 0.18 divided by 2 and multiplied by 3 (because of the coeefficients). We only have 0.0015 mol of Barium Hydroxide though, which tells us that this is the limiting reagent.

Since Barium Hydroxide is our LR, we will use 0.0015 mol of Barium Hydroxide mixed with excess phosphoric acid. We need to compare our molar ratios for Barium Hydroxide and Barium phosphate. We see that for 3 moles of Barium Hydroxide, we create only 1 mol of Barium phosphate. This means if we use 0.0015 mol of Barium Hydroxide, we create 0.015 / 3 mols of Barium phosphate = 0.005 mol

0.005 mol of Barium phosphate = x grams / 602 g FW

x = 3 grams

4)

pH = - log H+

pH = - log (3.88x10^-12)

pH = 11.41

pH + pOH = 14

11.41 + pOH = 14

pOH = 2.59

pH = - log H+

11.54 = - log H+

-11.54 = log H+

10^ -11.54 = H+

2.88x10^-12 = H+

H+ = OH- on both of these examples

and pOH would be 14-11.54 which equals 2.46

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