Question

How many milliliters of 0.25 N NaOH would be needed to completely neutralize

100 ml of this acid?​

Answer 1

Answer:

N1V1 = N2V2 works fine if you have values for three of the four variables. You can then solve for the fourth variable. As you correctly observe here, that approach won’t work.

But the reason N1V1 = N2V2 works … ahhhh … that’s where we can make some headway.

Normality is equivalents/L, so

N×V=equivL×L=equiv

So what we’re really saying with N1V1 = N2V2 is …

equivalents of the first reagent = equivalents of the second reagent

One way to find equivalents is by multiplying N × V if we have a solution of known normality and volume.

But if the reactant is not a solution of known normality and volume, we have to use other methods.

In this problem, the only thing we know about the quantity of our NaOH reactant is its mass. Can we convert that to equivalents? We can if we know the equivalent mass of NaOH. Since it reacts 1:1 with a monoprotic acid, equivalent mass = molar mass.

equiv NaOH=5 g×1 equiv40.00 g=0.125 equiv NaOH

So now we can say equiv HCl = equiv NaOH, or

NHCl×VHCl=0.125

0.25 equiv/L×VHCl=0.125 equiv

VHCl=0.5 L=500 mL