What volume of a 0.442 m naoh solution is needed to neutralize 65.0 ml of a 0.296 m solution of hno3?

The balanced equation for the reaction between NaOH and HNO₃ is
NaOH(aq) + HNO₃(aq) → NaNO₃(aq) + H₂O(l)

Molarity (M) = number of moles (mol) / Volume of the solution (L)

Molarity of HNO₃          = 0.296 M
Volume of the solution  = 65.0 mL = 65.0 x 10⁻³ L
Moles of HNO₃             = Molarity x Volume of the solution
                                     = 0.296 M x 65.0 x 10⁻³ L                       
                                     = 19.24 x 10⁻³ mol

Stoichiometric ratio between NaOH and HNO₃ is 1 : 1.
Hence,
   moles of NaOH = moles of HNO₃
                             = 19.24 x 10⁻³ mol

Molarity of NaOH = 0.442 M
Volume of NaOH = moles of NaOH / Molarity 
                            = 19.24 x 10⁻³ mol / 0.442 M
                            = 0.0435 L
                            = 43.5 mL

Hence, the volume of needed NaOH solution is 43.5 mL.

 The  volume of a 0.442 M NaOH  solution   needed  to neutralize  65.0 Ml  of  a 0.296 M  solution  of HNo3 is  43.5  ML

    Calculation
write the  equation  for neutralization
 NaOH   +HNO3  → NaNO3 +H2O

  find the     moles    HNO3  used

moles =molarity  x volume in liters
volume in  liters =65/1000 =0.065 liters
molarity =0.296 m

moles = 0.065   x 0.296 = 0.01924  moles

by  use of mole ratio  between    NaOH  to  HNO3   which is 1:1  the  moles of NaOH  is  also  0.01924  moles

The  volume of    NaOH  is =  moles/  molarity
molarity  =0.442 M

      0.01924/0.442  = 0.0435 L  or  0.0435  x1000  = 43.5  Ml