Question

What is the molarity of a naoh solution if 28.2 ml of a 0.355 m h2so4 solution is required to neutralize a 25.0-ml sample of the naoh solution?what is the molarity of a naoh solution if 28.2 ml of a 0.355 m h2so4 solution is required to neutralize a 25.0-ml sample of the naoh solution?0.8011250.3150.6290.400?

Answer 1

The  molarity  of  NaOH  solution  is  calculated  as   follows
step  one ;calculate  the  number  of  moles  of  H2SO4

moles  of  H2SO4=  molarity  x  volume/1000
                             =28.2   x0.355/1000=  0.01  moles  of  H2SO4

2NaOH  +H2SO4  ---->Na2SO4  +2H2O
from  the    equation   above  the   ratio  of  NaOH  to  H2SO4  is  2:1  therefore  the  moles  of   NaOH  is  =  2  x  0.01=0.02  moles

molarity  of   NaOH=  moles  of  NaOh/volume  x  1000

that  is  0.02/25   x1000=  0.8 M

Answer 2

Answer;

Molarity of NaOH is 0.80 M

Explanation;

The balanced equation for the reaction is;

2NaOH(aq) + H2SO4(aq = NaSO4(aq) +2 H2O (l)

Moles = concentration x volume  

thus; 0.355M x 0.0282L= 0.01 moles of H2SO4.

Using the mole ratio;

Moles of NaOH = Moles of H2SO4 ×2

                         = 0.02 Moles

Therefore; moles of NaOH = 0.02 moles

But; Concentration = moles / volume  

Thus; Concentration of NaOH = 0.02 / 0.025L

                                       = 0.8M