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?

M(HCl) * V(HCl) = M(NaOH) * V(NaOH)

M(HCl) = 0.35
V(HCl) = 45mL
M(NaOH)= 0.35

now, solne for V(NaOH) by putting these values in the above equation.
M(HCl) * V(HCl) = M(NaOH) * V(NaOH)

0.35 * 45 = 0.35 * V(NaOH)

V(NaOH) = 45 mL

The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃  → 3 nicl₂ + 2 albarium and copper bromideba + cubr  →  no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al  → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn  → znso₄ + na