Question

How many grams of NaH2PO4 are needed to react with 39.97 mL of 0.265 M NaOH?

NaH2PO4(s) + 2 NaOH(aq) Na3PO4(aq) + 2 H2O(l)

Answer 1

0.636 g NaH₂PO₄

Step-by-step explanation:

To find the mass of NaH₂PO₄ required, you need to (1) find moles NaOH (via the molarity equation using the molarity and volume (L)), then (2) convert moles NaOH to moles NaH₂PO₄ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles to grams (via molar mass from periodic table). The final answer should have 3 sig figs to match the given value with the lowest amount of sig figs (0.265 M).

(Step 1)

39.97 mL / 1,000 = 0.03997 L

Molarity (M) = moles / volume (L)

0.265 M = moles / 0.03997 L

(0.265 M) x (0.03997 L) = moles

0.0106 = moles NaOH

(Steps 2 + 3)

1 NaH₂PO₄ (s) + 2 NaOH (aq) ---> Na₃PO₄ (aq) + 2 H₂O

Molar Mass (NaH₂PO₄):

22.99 g/mol + 2(1.008 g/mol) + 30.974 g/mol + 4(16.00 g/mol)

Molar Mass (NaH₂PO₄): 119.98 g/mol

0.0106 moles NaOH 1 mole NaH₂PO₄ 119.98 g
------------------------------- x ---------------------------- x ------------------ =
2 moles NaOH 1 mole

= 0.636 g NaH₂PO₄