In a titration of 0.35 m hcl and 0.35 m naoh, how much naoh should be added to 45.0 ml of hcl to completely neutralize the acid?

It would be the same amount. So, 45 ml of NaOH is required to be added to the 45 ml of HCI to neutralize the acid fully. Here is a brief calculation:

Firstly, here is your formula: M(HCI) x V(HCI) = M(NaOh) x V(NaOH) 
With the values put in: 0.35 x 45 = 0.35 x V(NaOH) 
= 45 ml. 
There is 45 ml of V(NaOH)

Let me know if you need anything else. :)

           - Dotz

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