Question

Enter your answer in the provided box. How many grams of NaH2PO4 are needed to react with 26.93 mL of 0.391 M NaOH? NaH2PO4(s) + 2NaOH(aq) → Na3PO4(aq) + 2H2O(l)

Answer 1

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

0.3910 = number of moles of NaOH / 0.02693 L

number of moles of NaOH = 0.3910 * 0.02693 = 0.01052 mole of NaOH

from the balanced equation we can say that

2 mole of NaOH requires 1 mole of NaH₂PO₄

so 0.010529 mole of NaOH will require

= 0.010529 mole of NaOH*(1 mole of NaH₂PO₄ / 2 mole of NaOH)

= 5.264×10^-3 mole of NaH₂PO₄

1 mole of NaH₂PO₄ = 119.98 g

so 5.264×10^-3  mole of NaH₂PO₄ = 0.6317 g

Therefore, the amount of NaH₂PO₄ needed will be 0.6317 g