A student performing this experiment mistakenly used 6.0 mL of 16 M HNO3 to dissolve 0.18g of solid copper instead of the 4.0mL described in the lab manual.

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 4.0mL described in the lab manual. What volume in mL of 6.0M NaOH are required to neutralize the excess acid. 

Answer 1

The excess acid would be 2 ml of 16 M HNO₃ assuming the 4.0 ml of acid was the exact amount needed for 0.18 g of Cu. You can do tge calculations to determine this. 0.18 g Cu / atomic mass Cu = moles Cu. Is this = to moles acid using 4 ml of of 16 M acid?

moles HNO₃ = 0.002 L x 16 mol/L = 0.032 moles in excess

(xL)(6.0 mol/L) = 0.032 mol

x = 0.00533 liters = 5.33 mls of NaOH