Acid/Base neutralization

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Answer -   Normality of HCl  is  1.546 N   and   Normality of NaOH = 1.58 N Explanation -At end point ,     Mole of HCl = Mole of NaOH=> Mole of CaCO3 in HCl solution   = Mole of NaOH         ( as CaCO3  act as acid  )=> Concentration of CaCO3  x Volume of HCl  in Liter =  Concentration of NaOH  x  Volume of  NaOH  in Liter=> Concentration of CaCO3  x 0.04985 L =  Concentration of NaOH  x  0.00632 L=> ( Mole of CaCO3 / Volume of HCl in Liter )  x 0.04985 L =  Concentration of NaOH  x  0.00632 L=> ( Mole of CaCO3 /  0.04985 L )  x 0.04985 L =  Concentration of NaOH  x  0.00632 L=>   Mole of CaCO3  =  Concentration of NaOH  x  0.00632 L=>   ( weight of CaCO3 / molar mass of CaCO3 )  =  Concentration of NaOH  x  0.00632 L=>   ( 1.000 g / 100.0869 g/mol )  =  Concentration of NaOH  x  0.00632 L=>   ( 0.00999 mol )  =  Concentration of NaOH  x  0.00632 L
=>   ( 0.00999 mol ) /  0.00632 L =  Concentration of NaOH  =>    1.58 mol / L =  Concentration of NaOH  =>    1.58 M =  Concentration of NaOH As 1 M  = 1 N  ,         so,      Normality of NaOH = 1.58 N        ( Molarity ( M ) , Normality ( N )  )        Mole of HCl = Mole of NaOH
=> Concentration of HCl  x Volume of HCl  in Liter =  Concentration of NaOH  x  Volume of  NaOH  in Liter

=> Concentration of HCl  x 0.050 L =  1.58 M  x  0.04895 L

=> Concentration of HCl   =  ( 1.58 M  x  0.04895 L ) / 0.050 L
=>  Concentration of HCl   = 1.546 M         As 1 M  = 1 N  ,         so,
      Normality of HCl = 1.546 N Cheers...Have a good day...

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Acid/Base neutralization

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