To determine the percentage of sodium chloride in the mixture of sodium hydroxide (NaOH) and sodium chloride (NaCl), we first need to analyze the neutralization reaction between sodium hydroxide and sulfuric acid (H₂SO₄).
The balanced reaction for the neutralization is:
\[ 2 , \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 , \text{H}_2\text{O} \]
From the equation, we see that 2 moles of NaOH react with 1 mole of H₂SO₄.
Step 1: Calculate the moles of H₂SO₄ used.
Given that the concentration of H₂SO₄ is 0.25 M and the volume used is 20 cm³ (which is 0.020 L), we can find the moles of H₂SO₄ used:
\[ \text{Moles of H}_2\text{SO}_4 = \text{Concentration} \times \text{Volume} = 0.25 , \text{mol/L} \times 0.020 , \text{L} = 0.005 , \text{moles} \]
Step 2: Determine the moles of NaOH that reacted.
From the balanced equation, 2 moles of NaOH react with 1 mole of H₂SO₄. Therefore, the moles of NaOH that reacted can be calculated using the stoichiometric ratio:
\[ \text{Moles of NaOH} = 2 \times \text{Moles of H}_2\text{SO}_4 = 2 \times 0.005 , \text{moles} = 0.01 , \text{moles} \]
Step 3: Find the total amount of NaOH in the 50 cm³ of the original solution.
The 50 cm³ of solution represents \( \frac{50}{1000} = 0.050 , \text{L} \) of the total 1 L solution. Thus, the concentration of NaOH in the entire solution can be calculated:
\[ \text{Total moles of NaOH in 1 L} = \text{Moles of NaOH in 50 cm}^3 \times \frac{1 , \text{L}}{0.050 , \text{L}} = 0.01 , \text{moles} \times 20 = 0.2 , \text{moles} \]
Step 4: Calculate the mass of NaOH.
To convert moles of NaOH to grams, we need its molar mass:
Molar mass of NaOH = \( 23 + 16 + 1 = 40 , \text{g/mol} \)
\[ \text{Mass of NaOH} = \text{Moles} \times \text{Molar mass} = 0.2 , \text{moles} \times 40 , \text{g/mol} = 8 , \text{g} \]
Step 5: Determine the mass of the sodium chloride in the mixture.
We are given that the mixture of NaOH and NaCl weighs 8.8 g. If there are 8 g of NaOH in the mixture, we can find the mass of NaCl:
\[ \text{Mass of NaCl} = \text{Total mass of mixture} - \text{Mass of NaOH} = 8.8 , \text{g} - 8.0 , \text{g} = 0.8 , \text{g} \]
Step 6: Calculate the percentage of sodium chloride in the mixture.
Now we can find the percentage of NaCl in the mixture:
\[ \text{Percentage of NaCl} = \left( \frac{\text{Mass of NaCl}}{\text{Total mass of mixture}} \right) \times 100 = \left( \frac{0.8 , \text{g}}{8.8 , \text{g}} \right) \times 100 \approx 9.09% \]
Conclusion
The percentage of sodium chloride in the mixture is approximately 9.09%.
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