At 25 degree C, a student adds 0.400 g of solid phosphoric acid to a beaker containing 500.0ml of water. After the solid is completely dissolved the student adds 20.0ml of a 0.995M sodium hydroxide. What is the pOH of the solution in the beaker?
Interesting because at 25C H3PO4 is not solid. It freezes about 17 C.
mols H3PO4 = 0.4g/molar mass H3PO4 = estimated 0.004
mols NaOH added = M x L = estimated 0.02
You need to clean up the numbers before continuing.
......H3PO4 + 3NaOH ==> Na3PO4 + 3H2O
Starting with 0.02 mols NaOH, how much Na3PO4 will be formed. That's 0.02/3 = 0.00667 mols Na3PO4.
Starting with 0.004 mols H3PO4, how much Na3PO4 will form? That's 0.004. In limiting regent problems the smaller number is ALWAYS the correct amount and the reagent producing that number is the limiting reagent. Therefore, the H3PO4 is the limiting reagent.
.....H3PO4 + 3NaOH ==> Na3PO4 + 3H2O
I...0.004......0.........0........0
add.........0.02....................
C..-0.004..-0.012.......+0.004...+0.004
E.....0.....0.008.......0.0004...0.004
So you have a solution of about 0.008 mols NaOH in 520 mL which makes (OH^-) about 0.008/0.520 and convert to pOH. The PO4^3- formed will hydrolyze and contribute some OH^- but I doubt that will be significant.
To find the pOH of the solution in the beaker, we need to calculate the concentration of hydroxide ions (OH-) in the solution.
First, let's find the number of moles of phosphoric acid added to the beaker. We can use the formula:
moles = mass / molar mass
The molar mass of phosphoric acid (H3PO4) can be calculated by adding up the atomic masses of its elements:
(3 * atomic mass of hydrogen) + (1 * atomic mass of phosphorus) + (4 * atomic mass of oxygen).
Using the atomic masses from the periodic table, we get:
(3 * 1.008 g/mol) + (1 * 30.97 g/mol) + (4 * 16.00 g/mol) = 98.0 g/mol.
So, the number of moles of phosphoric acid is:
moles = 0.400 g / 98.0 g/mol = 0.00408 mol.
Next, let's calculate the number of moles of sodium hydroxide (NaOH) added to the beaker. We can use the formula:
moles = volume * molarity
Given that the volume is 20.0 ml and the molarity is 0.995 M, we can substitute these values into the formula:
moles = 20.0 ml * (0.995 mol/L / 1000 ml/L) = 0.0199 mol.
Since sodium hydroxide is a strong base, it completely dissociates in water, yielding one hydroxide ion for every sodium hydroxide molecule. Therefore, the concentration of hydroxide ions (OH-) in the solution is also 0.0199 mol.
Now that we know the concentration of hydroxide ions, we can calculate the pOH using the formula:
pOH = -log[OH-].
Taking the negative logarithm of 0.0199 gives us the pOH:
pOH = -log(0.0199) ≈ 1.70.
Therefore, the pOH of the solution in the beaker is approximately 1.70.