To determine how many grams of NaH₂PO₄ are needed to react with a given amount of NaOH, we first need to examine the chemical reaction between NaH₂PO₄ (sodium dihydrogen phosphate) and NaOH (sodium hydroxide).
The balanced chemical equation for the reaction is:
\[ \text{NaH}_2\text{PO}_4 + \text{NaOH} \rightarrow \text{Na}_2\text{HPO}_4 + \text{H}_2\text{O} \]
From the equation, we see that the reaction goes in a 1:1 mole ratio; one mole of NaH₂PO₄ reacts with one mole of NaOH.
Step 1: Calculate moles of NaOH
First, we need to calculate the number of moles of NaOH present in 20.78 mL of a 0.357 M solution.
Using the formula for moles:
\[ \text{Moles of NaOH} = \text{Concentration (M)} \times \text{Volume (L)} \]
Convert the volume from mL to L:
\[ 20.78 , \text{mL} = 0.02078 , \text{L} \]
Now, calculate the moles of NaOH:
\[ \text{Moles of NaOH} = 0.357 , \text{mol/L} \times 0.02078 , \text{L} = 0.00741 , \text{mol} \]
Step 2: Calculate moles of NaH₂PO₄ required
Since the reaction has a 1:1 ratio, the moles of NaH₂PO₄ required will be equal to the moles of NaOH.
\[ \text{Moles of NaH}_2\text{PO}_4 = 0.00741 , \text{mol} \]
Step 3: Calculate the mass of NaH₂PO₄ needed
Next, we will need the molar mass of NaH₂PO₄. The molar mass can be calculated as follows:
- Sodium (Na): 22.99 g/mol
- Hydrogen (H): 1.01 g/mol
- Phosphorus (P): 30.97 g/mol
- Oxygen (O): 16.00 g/mol
Calculating the molar mass:
\[ \text{Molar mass of NaH}_2\text{PO}_4 = 22.99 + (2 \times 1.01) + 30.97 + (4 \times 16.00) \] \[ = 22.99 + 2.02 + 30.97 + 64.00 \] \[ = 119.98 , \text{g/mol} \]
Step 4: Calculate the mass of NaH₂PO₄
Now, we can use the moles of NaH₂PO₄ to find the mass required:
\[ \text{Mass of NaH}_2\text{PO}_4 = \text{Moles} \times \text{Molar mass} \] \[ = 0.00741 , \text{mol} \times 119.98 , \text{g/mol} \approx 0.890 , \text{g} \]
Conclusion
Therefore, approximately 0.890 grams of NaH₂PO₄ are needed to react with 20.78 mL of 0.357 M NaOH.