Hello!
The reaction for the dissolving of solid copper with HNO₃ is the following:
Cu (s) + 4HNO₃ (aq) → Cu(NO₃)₂ (aq) + 2NO₂ (g) + 2H₂O (l)
The required mL to neutralize 0,18 g of solid copper are calculated using the following conversion factor:
[tex]0,18g Cu* \frac{1 mol Cu}{63,55 g Cu}* \frac{4 mol HNO3}{1 mol Cu}* \frac{1 L}{16 mol HNO3} * \frac{1000 mL}{1 L}= 0,71 mL[/tex]
Now we subtract this value to the volume of HNO3 added by the student:
[tex]Excess HNO3= VHNO3_{std} -VHNO3_{req}=6 mL-0,71 mL \\ =5,29mL[/tex]
To finish, we calculate the volume of NaOH 6M required to neutralize this amount of 16M acid in excess. The reaction is the following:
HNO₃ + NaOH → NaNO₃ + H₂O
To calculate this value, we use the following conversion factor:
[tex]V NaOH req=5,29mL_{HNO3}* \frac{1L}{1000mL} * \frac{16 mol HNO3}{1 L HNO3}* \frac{1 mol NaOH}{1 mol HNO3} \\ * \frac{1 L NaOH}{6 mol NaOH}* \frac{1000 mL}{1L}=14,11 mL NaOH
[/tex]
So, you'll need 14,11 mL of 6M NaOH to neutralize the excess acid
