Question

Determine the molarity of a NaOH solution when each of the following amounts of acid neutralizes 25.0 mL of the NaOH solution.

(a) 10.0 mL of 0.250 M HNO3	M
(b) 20.0 mL of 0.500 M H2SO4	M
(c) 25.0 mL of 1.00 M HCl	M
(d) 5.00 mL of 0.100 M H3PO4	M

Show Calculations.

Answer 1

a. HNO3 + NaOH----> NaNO3 + H2O
1 mole 1mole
HNO3
Molarity M1   = 0.25M
volume   V1   = 10ml
no of moles n1= 1 mole
NaOH
molarity   M2 =
volume     V2 = 25ml

no of molesn2 = 1
M1V1/n1 = M2V2/n2
M2      = M1V1*n2/n1V2
        = 0.25*10*1/1*25
        = 0.1M
Molarity of NaOH is 0.1M
b.H2SO4 + 2NaOH ---> Na2SO4 + H2O
1 mole   2 moles
H2SO4
molarity   M1 = 0.5M
volume     V1 = 20ml
no of moles n1 = 1 moles
NaOH
molarity   M2   =
volume     V2   = 25ml
no of moles n2 = 2 moles
M1V1/n1         =M2V2/n2
M2             = M1V1n2/n1V2
               = 0.5*20*2/1*25
               = 0.8M
molarity of NaOH = 0.8M
c.HCl   +   NaOH ----> NaCl + H2O
1 mole    1 mole
Hcl
molarity   M1 = 1M
volume     V1 = 25ml
no of moles n1 = 1 moles
NaOH
molarity   M2   =
volume     V2   = 25ml
no of moles n2 = 1 moles
M1V1/n1         =M2V2/n2
M2             = M1V1n2/n1V2
               = 1*25*1/1*25
               = 1M
molarity of NaOH = 1M
d.H3PO4 + 3 NaOH ----> Na3PO4 + 3H2O
1 mole   3 moles

H3PO4
molarity   M1 = 0.1M
volume     V1 = 5ml
no of moles n1 = 1 moles
NaOH
molarity   M2   =
volume     V2   = 25ml
no of moles n2 = 3 moles
M1V1/n1         =M2V2/n2
M2             = M1V1n2/n1V2
               = 0.1*5*3/1*25
               = 0.06M
molarity of NaOH is 0.06M