A 10.00mL solution of 0.2504M H3PO4 was titrated with standard 0.1010M NaOH. What volume of the bases was required to neutralized the acid

Answer 1

3 moles of NaOH required to neutralize 1 mole of H₃PO₄

no.of moles of H₃PO₄ = (0.2504 mol/L)(1 L/1000 mL)(10.00 mL) = 0.002504 mol

no.of moles of NaOH required to neutralize the H₃PO₄ = 3 X 0.002504 = 0.007512 mol

mL 0.1010 M NOH required to neutralize H₃PO₄ = [(0.007512 mol) / (0.1010 mol/L)] X 1000 mL/L = 74.38 mL

Answer 2

Write the balanced equation...

H₃PO₄ + 3NaOH ==> 3H₂O + Na₃PO₄

---Now, we calculate the moles of H3PO4 that are present:

10.00 ml x 1 L/1000 ml x 0.2504 moles/L = 0.002504 moles H₃PO₄ present

---Next, we find the moles of NaOH needed to neutralize/react with this many moles of H₃PO₄:

From the balanced equation, we see it takes THREE moles NaOH for each ONE mole H₃PO₄, thus..

0.002504 moles H₃PO₄ x 3 moles NaOH/1 mol H₃PO₄ = 0.007512 moles NaOH needed

---Finally, we calculate the volume of NaOH needed:

0.1010 mol/L (x L) = 0.007512 moles

x = 0.07438 L x 1000 ml/L = 74.38 mls NaOH