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 neutralize the acid, determine the concentration of the Ba(OH)2 solution used
moles acid=.175*75/98=.133 moles
you will need 3/2 *.133 moles Ba(OH)2 to neutralize...(write the balanced chem eq)
MolesBa(OH2)=Molarity*.215
Molarity=.133*1.5/.215=.93 check the math
To determine the concentration of the Ba(OH)2 solution used, we can use the concept of stoichiometry and the given information.
First, let's calculate the moles of H3PO4 in the 75.0g sample.
The percentage by mass of H3PO4 tells us that there are 17.5g of H3PO4 in every 100g of the liquid sample.
Moles of H3PO4 = Mass of H3PO4 / Molar mass of H3PO4
= 17.5g / 98.0g/mol
≈ 0.1786 mol
From the balanced chemical equation between H3PO4 and Ba(OH)2:
2H3PO4 + 3Ba(OH)2 -> Ba3(PO4)2 + 6H2O
We can see that 2 moles of H3PO4 react with 3 moles of Ba(OH)2.
Using stoichiometry, we can calculate the moles of Ba(OH)2 used to neutralize the H3PO4 in the sample:
Moles of Ba(OH)2 = (0.1786 mol H3PO4) x (3 mol Ba(OH)2 / 2 mol H3PO4)
≈ 0.268 mol
Now, we need to determine the volume of the Ba(OH)2 solution used. Given that 215.0mL of Ba(OH)2 is needed to completely neutralize the acid:
Volume of Ba(OH)2 solution used = 215.0 mL = 0.215 L
Finally, we can calculate the concentration of the Ba(OH)2 solution:
Concentration (Molarity) = Moles of solute / Volume of solution in liters
= (0.268 mol) / (0.215 L)
≈ 1.247 M
Therefore, the concentration of the Ba(OH)2 solution used is approximately 1.247 M.