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
To find the concentration of the Ba(OH)2 solution used, we need to determine the number of moles of Ba(OH)2 used and then divide it by the volume of the solution in liters.
First, let's find the number of moles of H3PO4 in the 75.0g sample.
To do this, we'll use the given percentage by mass of H3PO4, which is 17.5%.
Step 1: Calculate the mass of H3PO4 in the sample:
mass of H3PO4 = 17.5% * 75.0g
mass of H3PO4 = 0.175 * 75.0g
mass of H3PO4 = 13.125g
Step 2: Convert the mass of H3PO4 to moles of H3PO4 using its molar mass:
moles of H3PO4 = mass of H3PO4 / molar mass of H3PO4
moles of H3PO4 = 13.125g / 98.0g/mol
moles of H3PO4 = 0.1340 mol
Now, let's determine the number of moles of Ba(OH)2 used to neutralize the H3PO4.
The balanced equation for the reaction between H3PO4 and Ba(OH)2 is:
2H3PO4 + 3Ba(OH)2 -> Ba3(PO4)2 + 6H2O
From the equation, we can see that 2 moles of H3PO4 react with 3 moles of Ba(OH)2.
Step 3: Calculate the moles of Ba(OH)2 used:
moles of Ba(OH)2 = (moles of H3PO4 * 3) / 2
moles of Ba(OH)2 = (0.1340 mol * 3) / 2
moles of Ba(OH)2 = 0.201 mol
Next, we need to convert the volume of Ba(OH)2 solution from milliliters to liters.
The given volume is 215.0 mL.
Step 4: Convert the volume of Ba(OH)2 solution to liters:
volume of Ba(OH)2 solution = 215.0 mL * (1 L / 1000 mL)
volume of Ba(OH)2 solution = 0.215 L
Finally, we can calculate the concentration of the Ba(OH)2 solution used by dividing the moles of Ba(OH)2 by the volume of the solution in liters.
Step 5: Calculate the concentration of the Ba(OH)2 solution:
concentration = moles of Ba(OH)2 / volume of solution
concentration = 0.201 mol / 0.215 L
concentration = 0.935 mol/L
Therefore, the concentration of the Ba(OH)2 solution used is 0.935 mol/L.