A student is asked to standardize a solution of barium hydroxide. He weighs out 0.986 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid).
It requires 27.8 mL of barium hydroxide to reach the endpoint.
A. What is the molarity of the barium hydroxide solution(in M)
i got it, it's 0.111M!
Your data allows three places in the answer and you have only two.
To determine the molarity of the barium hydroxide solution, we need to use the stoichiometry of the reaction between the barium hydroxide and the potassium hydrogen phthalate. The balanced equation for the reaction is:
2KHC8H4O4 + Ba(OH)2 → K2Ba(C8H4O4)2 + 2H2O
In this reaction, one mole of barium hydroxide (Ba(OH)2) reacts with two moles of potassium hydrogen phthalate (KHC8H4O4) to form one mole of the diacid salt (K2Ba(C8H4O4)2) and two moles of water (H2O).
First, we need to calculate the number of moles of potassium hydrogen phthalate used:
mol KHC8H4O4 = mass of KHC8H4O4 / molar mass of KHC8H4O4
The molar mass of KHC8H4O4 is calculated as follows:
(1 x molar mass of K) + (1 x molar mass of H) + (8 x molar mass of C) + (4 x molar mass of H) + (4 x molar mass of O)
Let's calculate the molar mass of KHC8H4O4:
(1 x 39.10 g/mol) + (1 x 1.01 g/mol) + (8 x 12.01 g/mol) + (4 x 1.01 g/mol) + (4 x 16.00 g/mol) = 204.23 g/mol
Now, calculate the number of moles of KHC8H4O4 used:
mol KHC8H4O4 = 0.986 g / 204.23 g/mol = 0.00483 mol
Next, we use the stoichiometry of the reaction to determine the number of moles of barium hydroxide used, since it reacts in a 1:2 ratio with the potassium hydrogen phthalate.
Since two moles of KHC8H4O4 react with one mole of Ba(OH)2, we have:
moles Ba(OH)2 = 0.00483 mol KHC8H4O4 x (1 mol Ba(OH)2 / 2 mol KHC8H4O4) = 0.00241 mol Ba(OH)2
Finally, we can calculate the molarity (M) of the barium hydroxide solution using the definition of molarity:
Molarity (M) = moles Ba(OH)2 / volume (in liters) of solution
Since the volume is given in milliliters, we need to convert it to liters:
Volume of solution = 27.8 mL = 27.8 mL x (1 L / 1000 mL) = 0.0278 L
Now we can calculate the molarity:
Molarity (M) = 0.00241 mol Ba(OH)2 / 0.0278 L = 0.0868 M
Therefore, the molarity of the barium hydroxide solution is 0.0868 M.