A 75.0g sample of liquid contains 17.5% by mass of H3PO4 (molar mass=98.0g/mol). If 215.0mL of Ba(OH)2 is needed to completely neutralized the acid, determine the concentration (or molarity) of the Ba(OH)2 solution used.
2H3PO4 + 3Ba(OH)2 ==> Ba3(PO4)2 + 6H2O
mass H3PO4 in the sample = 75.0 x 0.175 = approx 13 but this is just a close estimate. You need to redo all of these calculations more accurately.
mols H3PO4 = appox 13/98 = 0.13
Using the coefficients in the balanced equation, convert mols H3PO4 used to mols Ba(OH)2 needed. 0.13 mols H3PO4 x (3 mols Ba(OH)2/2 mols H3PO4) = approx 0.2 mols Ba(OH)2
[Ba(OH)2] = mols/L = approx 0.2/0.215 = ?
To determine the concentration or molarity of the Ba(OH)2 solution used, we need to first calculate the number of moles of H3PO4 in the 75.0g sample and then use this information to find the molarity of the Ba(OH)2 solution.
Step 1: Calculate the number of moles of H3PO4.
To do this, we need to determine the mass of H3PO4 in the 75.0g sample.
Mass of H3PO4 = 17.5/100 * 75.0g = 13.125g
Next, we convert the mass of H3PO4 to moles using the molar mass of H3PO4.
Number of moles of H3PO4 = Mass of H3PO4 / Molar mass of H3PO4
Number of moles of H3PO4 = 13.125g / 98.0g/mol = 0.134mol
Step 2: Determine the molarity of the Ba(OH)2 solution.
We know that 215.0mL of the Ba(OH)2 solution is needed to completely neutralize the 0.134mol of H3PO4.
Molarity (M) is defined as moles of solute divided by the volume of the solution in liters.
Molarity (M) = moles of solute / volume of solution in liters
In this case, the solute is Ba(OH)2 and the volume of the solution is given as 215.0mL, so we need to convert mL to L.
Volume of solution = 215.0mL / 1000 = 0.215L
Now, we can calculate the molarity of the Ba(OH)2 solution.
Molarity of Ba(OH)2 = moles of Ba(OH)2 / volume of solution in liters
Since the balanced chemical equation between H3PO4 and Ba(OH)2 is 3H3PO4 + 6Ba(OH)2 → Ba3(PO4)2 + 6H2O, we know that 3 moles of H3PO4 react with 6 moles of Ba(OH)2. This means 0.134mol of H3PO4 will react with (6/3) * 0.134mol of Ba(OH)2.
Moles of Ba(OH)2 = (6/3) * 0.134mol = 0.268mol
Molarity of Ba(OH)2 = 0.268mol / 0.215L ≈ 1.25M
Therefore, the concentration or molarity of the Ba(OH)2 solution used is approximately 1.25M.