In the Cu-Zn alloy analysis, the sample is dissolved in excess HNO3, excess acid is boiled to adjust pH, and titrated with Na2S2O3. The Cu content is determined as approximately 2.31%.
1. **Reactions:**
a. Dissolving the alloy in excess HNO3:
\[Cu-Zn + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + Zn(NO_3)_2 + 4H_2O\]
b. Boiling excess acid and adjusting pH:
\[4HNO_3 \rightarrow 4NO_2 + 2H_2O + O_2\]
\[Cu(NO_3)_2 + 2H_2O \rightarrow Cu(OH)_2 + 2HNO_3\]
c. Titration with Na2S2O3:
\[2Cu^2+ + 5S_2O_3^{2-} + 2H_2O \rightarrow 2CuS + 5SO_4^{2-} + 4H^+\]
2. **Calculation:**
Moles of Na2S2O3 reacted = \(\frac{15.0 \, \text{mL} \times 0.1 \, \text{mol/L}}{1000 \, \text{mL/L}} = 0.0015 \, \text{mol}\)
Since 1 mole of Na2S2O3 reacts with 1 mole of Cu^2+, moles of Cu^2+ = 0.0015 mol
Since the reaction ratio is 2:1 (Cu^2+:Cu), moles of Cu in the alloy = \(0.0015 \times \frac{1}{2} = 0.00075 \, \text{mol}\)
Mass of Cu = \(0.00075 \, \text{mol} \times 63.55 \, \text{g/mol} = 0.0476625 \, \text{g}\)
**Cu content in the alloy** = \(\frac{0.0476625 \, \text{g}}{2.068 \, \text{g}} \times 100 \% \approx 2.31 \%\)
The probable question may be:
To determine the Cu content in Cu-Zn alloy, one does the following: Draw
Completely dissolve 2.068g of Cu-Zn alloy sample in excess acid HNO3, obtaining a solution
solution X. Boil excess acid, adjust to pH 3 to obtain 100mL of solution Y. Take
10mL of solution Y, add excess KI, then titrate the resulting solution with the solution
0.1M Na2S2O3 shows 15.0 mL. Write the equations for the reactions that occur. Calculate
Cu content in the above alloy sample.