Question

A Student 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

Answer 1

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:


`0,18g Cu* (1 mol Cu)/(63,55 g Cu)* (4 mol HNO3)/(1 mol Cu)* (1 L)/(16 mol HNO3) * (1000 mL)/(1 L)= 0,71 mL`

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


`Excess HNO3= VHNO3_(std) -VHNO3_(req)=6 mL-0,71 mL \\ =5,29mL`

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:


`V NaOH req=5,29mL_(HNO3)* (1L)/(1000mL) * (16 mol HNO3)/(1 L HNO3)* (1 mol NaOH)/(1 mol HNO3) \\ * (1 L NaOH)/(6 mol NaOH)* (1000 mL)/(1L)=14,11 mL NaOH`

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