We are given the following data box above and are asked to determine the Molarit

Question

We are given the following data box above and are asked to determine the Molarity of phosphoric acid

Additonal info, We pipeted 5.00 mL of 0.1M Phosphoric Acid into the beaker and added 40 mL of deionized water. The titrant used in this experiment was 0.0750 M NaOH. Let me know if more info is needed. The professor also mentioned that we should use the first equivalence point in order to find the molarity. I'm not really sure how to set it up, please show full work. Thanks help would be much appreciated!

Data Box Equivalence point 1 (mL titrant) Equivalence point 2 (mL titrant) 14 Half equivalence point 1 pH 2.47 Half equivalence point 2 pH pKa 1 2.47 pKa 2 Molarity of phosphoric acid (mol/L) 747272 14141 2525 tt nn aa rr ttll tt LL-2 rmtt 1200 pp tt nnee 00CC ppnn ee ccaa- nn>v eeuu Baaqq tuu aiiffaao aqqaall DEEHHppM

Explanation / Answer

Apply

mmol of base = mmol of acid (H+)

mmol of base in 1st point = MV = 0.075 * 7 = 0.525 mmol of NaOH used for first point

so

mmol of H+ = 0.525 mmol

Therefore

there is 1 first proton per 1 mol of H3PO4

so

mmol of H3PO4 = 0.525 mmol

Vtotal = 5+ 40 = 45

M = mmol/mL = 0.525/45 = 0.01166666667 mol per liter

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.