I'm sorry, but I'm not sure exactly what you're saying here.
Okay, bear with me. Had the original poster given actual reaction conditions, I could get this right. Let us assume we need to nitrate 1 mole and we used a 50% excess of 68% nitric acid.
Start: 1.5 mol x 63 g = 94.5 g HNO3
139 g x 68% = 94.5 g HNO3
139 g - 94.5 g = 44.5 g H2O
End:
94.5 g HNO3 - 63 g HNO3 = 31.5 g HNO3
44.5 g H2O + 18 g H2O = 62.5 g H2O
31.5 g HNO3/(31.5 g + 62.5 g) = 0.33
Without having done anything, will 62% nitric acid work in this reaction? Well, the concentration range for my hypothetical reaction gives the product with the nitric acid concentration falling from 68% down to 33%. So, I would say yes, 62% should work as well. If we knew the actual stoichiometry of the proposed reaction, we could calculate that final concentration.
If a smaller excess of nitric acid were used, then the concentration would fall to an even lower value. Since the reaction depletes the nitric acid, then a higher rate can be obtained by simply adding more nitric acid. Presumably, if one were forced to start with a lower concentration of nitric acid, one could increase the amount of nitric acid to obtain the same final concentration.