Question

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

Answer 1

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

Answer 2

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.