Astudent performing this experiment mistakenly used 6.0 ml of 16 m hno3 to dissolve 0.18g of solid copper , instead of the 4.0 ml described in lab manual. what volume of 6 m naoh are required to neutralized the excess acid

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:



Now we subtract this value to the volume of HNO3 added by the student:



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:



So, you'll need 14,11 mL of 6M NaOH to neutralize the excess acid