Question

How many grams of NaH2PO4 are needed to react with 30.66 mL of 0.260 M NaOH? NaH2PO4(s) + 2 NaOH(aq) → Na3PO4(aq) + 2 H2O(l)

Answer 1

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

0.260 = number of moles of NaOH / 0.03066 L

number of moles of NaOH = 0.260 * 0.03066 = 0.00797 mole of NaOH

from the balanced equation we can say that

2 mole of NaOH requires 1 mole of NaH₂PO₄

so 0.00797 mole of NaOH will require

= 0.00797 mole of NaOH*(1 mole of NaH₂PO₄ / 2 mole of NaOH)

= 0.003985 mole of NaH₂PO₄

1 mole of NaH₂PO₄ = 119.98 g

so 0.003985 mole of NaH₂PO₄ = 0.478 g

Therefore, the amount of NaH₂PO₄ needed will be 0.478 g