Question

If you use 1.98 ml (an excess) of HNO₃, what volume of 5.0M NaOH will you need to neutralize the solution after the copper is completely dissolved?

Answer 1

To neutralize 1.98 mL of an unspecified concentration of HNO₃, assuming it's 0.125M as indicated in the provided examples, you would need 49.5 µL of 5.0M NaOH, based on the 1:1 reaction stoichiometry between acid and base.

Step-by-step explanation:

To determine the volume of 5.0M NaOH needed to neutralize the solution containing the excess HNO3, we first need to consider the reaction between NaOH and HNO3, which is a typical acid-base neutralization reaction:

NaOH(aq) + HNO3(aq) → NaNO3(aq) + H2O(l)

Since this is a 1:1 reaction, the moles of NaOH needed will equal the moles of HNO3 present. Firstly, we convert the volume of HNO3 used to moles, using the relationship:

Molarity (M) = Moles/Vol(L)

1.98 mL is equivalent to 0.00198 L. If the initial concentration of HNO3 is not provided and assuming it is the same as the example given, which uses 0.125 M HNO3, we calculate the moles of HNO3:

Moles of HNO3 = 0.125 M × 0.00198 L = 0.0002475 mol

Then, to find the volume of the 5.0M NaOH needed to neutralize this amount of acid:

Vol(L) = Moles/Molarity

Vol(NaOH) = 0.0002475 mol / 5.0 M = 0.0000495 L or 49.5 µL

Therefore, 49.5 µL of 5.0M NaOH would be required to neutralize 1.98 mL of 0.125M HNO3.