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
How much H3PO4 is in the sample?
That's 75.0 x 0.175 = approx 13g but you need to do it more accurately. How many mols is that? That's about 13/98 = approx 0.13.
2H3PO4 + 3Ba(OH)2 ==> Ba3(PO4)2 + 6H2O
mols Ba(OH)2 = approx 0.13 mols H3PO4 x (3 mol Ba(OH)2/2 mol H3PO4) = about 0.13 x 3/2 = about 0.2 mols Ba(OH)2.
So M Ba(OH)2 = mols/L. YOu have mols and convert 215.0 mL to L.
Donloud
To determine the concentration of the Ba(OH)2 solution used, we need to calculate the number of moles of H3PO4 and Ba(OH)2 involved in the neutralization reaction.
First, let's calculate the number of moles of H3PO4:
1. Determine the mass of H3PO4 in the given liquid sample:
Mass of H3PO4 = mass of sample × percentage of H3PO4 / 100
= 75.0g × 17.5/100
= 13.125g
2. Convert the mass of H3PO4 to moles using its molar mass:
Moles of H3PO4 = Mass of H3PO4 / Molar mass of H3PO4
= 13.125g / 98.0g/mol
= 0.134 mols
Next, let's determine the moles of Ba(OH)2 required for neutralization:
The balanced chemical equation for the neutralization reaction between H3PO4 and Ba(OH)2 is:
2 H3PO4 + 3 Ba(OH)2 → Ba3(PO4)2 + 6 H2O
From the balanced equation, we can see that 2 moles of H3PO4 react with 3 moles of Ba(OH)2. Therefore, the mole ratio between H3PO4 and Ba(OH)2 is 2:3.
3. Calculate the moles of Ba(OH)2:
Moles of Ba(OH)2 = (Moles of H3PO4 × 3) / 2
= (0.134 mols × 3) / 2
= 0.201 mols
Finally, let's calculate the concentration of the Ba(OH)2 solution:
4. Determine the volume of the Ba(OH)2 solution in liters:
Volume of Ba(OH)2 solution = 215.0 mL = 215.0 cm^3 = 0.215 L
5. Calculate the concentration of the Ba(OH)2 solution:
Concentration (Molarity) = Moles of Ba(OH)2 / Volume of Ba(OH)2 solution
= 0.201 mols / 0.215 L
= 0.935 M
Therefore, the concentration of the Ba(OH)2 solution used is 0.935 M.