Question

At one point in the lab you will need to perform an acid-base neutralization reaction, using NaOH to neutralize HNO3. Calculate the required amount of NaOH needed for neutralization, knowing that nitric acid is 15.8 M and assuming 6 mL were used. In lab we will use a slight excess (+10%) of NaOH to ensure the acid is neutralized. HNO3 + NaOH à NaNO3 + H2O

Answer 1

no. of mole = molarity volume of solution in liter

6 ml = 0.006 liter

no. of mole of HNO₃ = 15.8 0.006 = 0.0948 mole

nutrilization reaction between HNO₃ and NaOH is

HNO₃ + NaOH NaNO₃ + H₂O

According to reaction both HNO₃ and NaOH required equimolar proportion for nuetrilization therefore to neutrilize 0.0948 mole of HNO₃ requied mole of NaOH = 0.0948 mole

molar mass of NaOH = 39.997 gm/mole then 0.0948 mole NaOH = 39.997 0.0948 = 3.79 gm

NaOH required 3.79 gm

volume of 10% NaOH solution in ml = mass of solute in gm 100 / % solution(m/v)

volume of 10% NaOH solution in ml = 3.79 100 / 10 = 37.9 ml

to neutrilize 15.8 M 6 ml HNO₃ required 10%(m/v) NaOH = 37.9 ml