Question

how many ml of 0.185 M NaOH are required to completely neutralize 25.0 ml of 0.135 M H3PO4

H3PO4 (aq) + 3NaOH (aq) --> Na3PO4(aq) + 3H20 (l)

Answer 1

Number of moles of H3PO4 = molarity * volume of solution in L

Number of moles of H3PO4 = 0.135 * 0.025 L = 0.00338 mole

from the balanced equation we can say that

1 mole of H3PO4 requires 3 mole of NaOH so

0.00338 mole of H3PO4 will require

= 0.00338 mole of H3PO4 *(3 mole of NaOH / 1 mole of H3PO4)

= 0.0101 mole of NaOH

molarity of NaOH = number of moles of NaOH / volume of solution in L

0.185 = 0.0101 / volume of solution in L

volume of solution in L = 0.0101 / 0.185 = 0.0546 L

1L = 1000 mL

0.0546 L = 54.6 mL

Therefore, the volume of NaOH required would be 54.6 mL