Question

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

Answer 1

The molarity of barium hydroxide solution is 0.109 M

Step-by-step 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

Answer 2

molarity of the barium hydroxide solution = 0.109 M

Step-by-step explanation:

monoprotic means it donates one proton or H+ ion for neutralization of 0.5 MOLE of Ba(OH)2

We first Write and balance the equation. If we call potassium hydrogen phthalate just KHP, then

Ba(OH)2 + 2KHP ==> 2H2O + BaP + K2P

Then Convert grams to moles. given mass of KHP =0.954g, molar mass of KHP= 204

moles of KHP = 0.954/molar mass

KHC8H4O4 = 0.954/204 = approximately 0.0047

Using the coefficients in the balanced equation, convert moles KHP to moles Ba(OH)2

0.0047moles KHP x (0.5)= 0.0047 x (1/2) = 0.00235 moles Ba(OH)2

M of Ba(OH)2 = moles Ba(OH)2/ volume of Ba(OH)2 in L

volume given =21.5 ml= 0.0215 L

M Ba(OH)2 = 0.00235/0.0215= approximately 0.109 M

so the molarity of the barium hydroxide solution = 0.109 M