If 250.0 mL of 5.52 M H3PO4 solution is added to 775 mL of 5.52 M NaOH solution,

Question

If 250.0 mL of 5.52 M H3PO4 solution is added to 775 mL of 5.52 M NaOH solution, the resulting solution will be ______ molar in Na3PO4 and _____ molar in ______.

Explanation / Answer

By balancing the equation: H3PO4 + 3NaOH --> Na3PO4 + 3H2O no of moles of H3PO4 give=volume(litre)*molarity=1.38 mole no of moles of NaOH given =volume*molarity=.775*5.52=4.278 by equation 1 mole of H3PO4 requires 3 mole of NaOH than given 1.38 mole of H3PO4 requires 3*1.38=4.14 mole of NaOH excess amount of Naoh = given-required=4.278-4.14=0.138 mole of NaOH 1 mole of H3PO4 produces 1 mole of Na3PO4 and 3 mole of H2O than 1.38 mole of H3PO4 will produce 1.38 mole of Na3PO4 and 3*1.38=4.14 mole of H2O total volume=250+775+no of mole of water*18 ml=1099.52ml=1.1 L molarity of NA3PO4=1.38/1.1=1.254545... molarity of NaOH=.138/1.1=0.1254545...

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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