Questions: How many grams of NaH2PO4 are needed to react with 64.67 mL of 0.919 M NaOH? Be sure your answer has the correct number of significant figures. g NaHPO4

Solution Steps

The reaction between sodium dihydrogen phosphate (\(\mathrm{NaH}_2\mathrm{PO}_4\)) and sodium hydroxide (\(\mathrm{NaOH}\)) can be represented as:

\[ \mathrm{NaH}_2\mathrm{PO}_4 + \mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{HPO}_4 + \mathrm{H}_2\mathrm{O} \]

This equation shows that one mole of \(\mathrm{NaH}_2\mathrm{PO}_4\) reacts with one mole of \(\mathrm{NaOH}\).

Given:

• Volume of \(\mathrm{NaOH}\) solution = 64.67 mL = 0.06467 L
• Molarity of \(\mathrm{NaOH}\) = 0.919 M

The number of moles of \(\mathrm{NaOH}\) is calculated using the formula:

\[ \text{moles of NaOH} = \text{Molarity} \times \text{Volume (L)} \]

\[ \text{moles of NaOH} = 0.919 \, \text{mol/L} \times 0.06467 \, \text{L} = 0.05943 \, \text{mol} \]

From the balanced equation, the mole ratio of \(\mathrm{NaH}_2\mathrm{PO}_4\) to \(\mathrm{NaOH}\) is 1:1. Therefore, the moles of \(\mathrm{NaH}_2\mathrm{PO}_4\) required are also 0.05943 mol.

The molar mass of \(\mathrm{NaH}_2\mathrm{PO}_4\) is calculated as follows:

• Na: 22.99 g/mol
• H: 1.01 g/mol (2 atoms)
• P: 30.97 g/mol
• O: 16.00 g/mol (4 atoms)

\[ \text{Molar mass of } \mathrm{NaH}_2\mathrm{PO}_4 = 22.99 + (2 \times 1.01) + 30.97 + (4 \times 16.00) = 119.98 \, \text{g/mol} \]

The mass of \(\mathrm{NaH}_2\mathrm{PO}_4\) required is:

\[ \text{mass} = \text{moles} \times \text{molar mass} \]

\[ \text{mass} = 0.05943 \, \text{mol} \times 119.98 \, \text{g/mol} = 7.134 \, \text{g} \]

Final Answer