A student is asked to standardize a solution of barium hydroxide. He weighs out 0.954 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 21.5 mL of barium hydroxide to reach the endpoint. A. What is the molarity of the barium hydroxide solution

The molarity of barium hydroxide solution is 0.109 M

Explanation:

Step 1: Data given

Mass of  potassium hydrogen phthalate (KHC8H4O4) = 0.954 grams

Molar mass of  potassium hydrogen phthalate = 204.22 g/mol

Volume of barium hydroxide (Ba(OH)2) = 21.5 mL = 0.0215 L

Step 2: The balanced equation

Ba(OH)2 + 2C8H5KO4 → Ba(C8H4KO4)2 + 2H2O

Step 3: Calculate moles potassium hydrogen phthalate

Moles potassium hydrogen phthalate = mass / molar mass

Moles potassium hydrogen phthalate = 0.954 grams / 204.22 g/mol

Moles potassium hydrogen phthalate = 0.00467 moles

Step 4: Calculate moles Ba(OH)2

For 1 mol Ba(OH)2 we need 2 moles C8H5KO4

For 0.00467 moles C8H5KO4 we need 0.00467/2 = 0.002335 moles

Step 5: Calculate molarity of Ba(OH)2

Molarity = moles / volume

Molarity Ba(OH)2 = 0.002335 moles / 0.0215 L

Molarity Ba(OH)2 = 0.109 M

The molarity of barium hydroxide solution is 0.109 M

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.