contestada

Automobile batteries use 3.0 m h2so4 as an electrolyte. how much 1.20 m naoh will be needed to neutralize 225 ml of battery acid? h2so4(aq) + 2naoh(aq) --> 2h2o(l) + na2so4(aq)

Respuesta :

kenmyna
H2SO4 (aq) +   2  NaOh (aq)  ------>2H2O   +  Na2SO4(aq)

V1M1 n2=  V2M2n1
  V=  volume
M=  concentration  in  mole  per   liter
n=  number  of  moles
V1=?
V2=  225  ml  
M1= 1.2  M 
M2=  3 m   
n1=2  moles
  V1   is  therefore  =  ( 225  x3  x2 )  /1.2  =    1125  ml

Answer : The volume of the [tex]NaOH[/tex] needed are, 1125 ml

Explanation :

Using neutralization law,

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1[/tex] = basicity of an acid = 2

[tex]n_2[/tex] = acidity of a base = 1

[tex]M_1[/tex] = concentration of [tex]H_2SO_4[/tex] = 3.0 M

[tex]M_2[/tex] = concentration of NaOH = 1.20 M

[tex]V_1[/tex] = volume of [tex]H_2SO_4[/tex] = 225 ml

[tex]V_2[/tex] = volume of NaOH = ?

Now put all the given values in the above law, we get the volume of the [tex]NaOH[/tex].

[tex]2\times 3.0M\times 225ml=1\times 1.20M\times V_2[/tex]

[tex]V_2=1125ml[/tex]

Therefore, the volume of the [tex]NaOH[/tex] needed are, 1125 ml

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