1) Using the 1st equivalence point from part a calculate the molarity of the phosphoric acid. Molarity of NaOH = .1523 M
Volume of phosphoric acid used 10.0mL
Volume at equivalence point 7.410 mL and 14.861 mL
2) Using the 2nd equivalence point from part a, calculate the molarity of the phosphoric acid. (remember that 2 H+ ions are being neutralized in this case.) How does this value compare to question one?
Any help would be appreciated! Thanks!
NaOH + H3PO4 ==> NaH2PO4 + H2O
mL x M NaOH = millimols.
That = mmols H3PO4
M H3PO4 = mmols H3PO4/mL H3PO4 = ?
2NaOH + H3PO4 ==>Na2HPO4
mL x M NaOH = 14.861 x 0.1538 = ?
mmols H3PO4 = ? x 1/2
M H3PO4 = mmols H3PO4/10 mL = ?
Thank you for your response! Having all of these numbers confused me. So for part a I'm multiplying 7.410 mL (first equivalence point) by .1523 M. I got 1.129 mmol H3PO4. Then divided it by 10.0mL to get a concentration of 0.113 M H3PO4.
I see I made a typo on molarity. The 0.1538 I used should be 0.1523.
You're right. 7.410 x 0.1523/10.0 = ?
14.810 x 0.1523 x 1/2 x 1/10.0 = ?
Thanks so much and the typo was no big deal I knew what you meant. :) you're a life saver!
To find the molarity of the phosphoric acid in question one and question two, you need to use the concept of stoichiometry and the volume of NaOH at the equivalence point.
1) First, let's calculate the number of moles of NaOH used in question one. To do this, we need to multiply the volume of NaOH used by its molarity.
Moles of NaOH = Volume of NaOH (L) * Molarity of NaOH (mol/L)
Since the volume of NaOH used is given as 10.0 mL (0.0100 L) and the molarity of NaOH is 0.1523 M, we can calculate the moles of NaOH used:
Moles of NaOH = 0.0100 L * 0.1523 mol/L = 0.001523 mol
Remember that the stoichiometric ratio between NaOH and phosphoric acid is 1:1. Therefore, the moles of NaOH used are equal to the moles of phosphoric acid used.
Now, let's calculate the molarity of the phosphoric acid in question one. To do this, divide the moles of phosphoric acid used by the volume of phosphoric acid used.
Molarity of phosphoric acid = Moles of phosphoric acid / Volume of phosphoric acid used (L)
Since the volume of phosphoric acid used is given as 10.0 mL (0.0100 L), we can calculate the molarity of phosphoric acid:
Molarity of phosphoric acid = 0.001523 mol / 0.0100 L = 0.1523 mol/L
2) For question two, we know that two H+ ions are being neutralized at the second equivalence point. This means that each mole of phosphoric acid will react with two moles of NaOH. Therefore, we need to multiply the moles of NaOH used by 2 to find the moles of phosphoric acid used.
Moles of phosphoric acid = Moles of NaOH * 2 = 0.001523 mol * 2 = 0.003046 mol
Now, let's calculate the molarity of the phosphoric acid in question two by dividing the moles of phosphoric acid used by the volume of phosphoric acid used (volume at the second equivalence point).
Molarity of phosphoric acid = Moles of phosphoric acid / Volume at equivalence point (L)
Since the volume at the second equivalence point is given as 14.861 mL (0.014861 L), we can calculate the molarity of phosphoric acid:
Molarity of phosphoric acid = 0.003046 mol / 0.014861 L = 0.205 mol/L
To compare the values obtained in question one and question two, the molarity of the phosphoric acid in question one is 0.1523 mol/L, while the molarity in question two is 0.205 mol/L. Therefore, the molarity of the phosphoric acid at the second equivalence point (question two) is higher than at the first equivalence point (question one).