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Question 1 of 3

What is the Normality of a solution resulting from the mixing of 0.5 moles of Sodium Hydroxide and 0.5 moles of Hydrochloric Acid if the final solution has a volume of 250 ml?
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N
The Normality of the solution is 2 N.
  1. The reaction involved is:NaOH + HCl ––––» NaCl + H2O
    Thus you are finding the Normality arising from the presence of Sodium Chloride.
  2. Find the mass of Sodium Chloride produced.
    The mixing of 0.5 moles of NaOH and 0.5 moles of HCl will produce 0.5 moles of NaCl.
    Mass of NaCl produced = Moles x RMM = 0.5 x 58.5 g = 29.25 g
  3. Find the number of equivalents of Sodium Chloride in 250 ml of solution.
    1 molecule of NaCl can replace 1 atom of Hydrogen.
    Thus the gram equivalent of NaCl is RMM/1 ie 58.5 g.
    Number of Equivalents =    Total Weight of NaCl    =    29.25   = 0.5 Eq
    NaCl Gram Equivalent       58.5
  4. 250 ml of solution contains 0.5 Equivalents. Thus:
    1000 ml of solution contains 1000/250 x 0.5 Eq = 2 Eq
    The concentration is 2 Eq/litre, giving a Normality of 2.

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Question 2 of 3

5.9 g of Succinic Acid and 4.0 g of Sodium Hydroxide are mixed together resulting in 300 ml of solution. What is the normality of the resulting solution? Answer to 4 decimal places.
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N
The Normality of the solution is 0.3333 N.
  1. The reaction involved is: (CH2COOH)2 + 2 NaOH ––––» (CH2COONa)2 + 2 H2O
    Thus you are finding the Normality arising from the presence of Sodium Succinate.
  2. Find the mass of Sodium Succinate produced.
    5.9/118 = 0.05 moles of Succinic Acid are mixed with 4.0/40 = 0.1 mole of NaOH.
    Thus 0.05 moles of Sodium Succinate are produced.
    Mass of Sodium Succinate produced = Moles x RMM = 0.05 x 162 g = 8.1 g
  3. Find the number of equivalents of Sodium Succinate in 300 ml of solution.
    1 molecule of Sodium Succinate can replace 2 atoms of Hydrogen.
    Thus the gram equivalent of Sodium Succinate is RMM/2 ie 81.0 g.
    Number of Equivalents =    Total Weight of Succinate    =   8.1   = 0.1 Eq
    Succinate Gram Equivalent     81.0
  4. 300 ml of solution contains 0.1 Equivalents. Thus:
    1000 ml of solution contains 1000/300 x 0.1 Eq = 0.33333 Eq = 0.3333 Eq to 4 dec pl
    The concentration is 0.3333 Eq/litre, giving a Normality of 0.3333.

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Question 3 of 3

What is the Normality of a solution resulting from the mixing of 2.45 g of Sulphuric Acid with 2.805 g of Potassium Hydroxide if the volume of the final solution is 650 ml. Answer to 5 decimal places.
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N
The Normality of the solution is 0.07692 N.
  1. The reaction involved is: H2SO4 + 2 KOH ––––» K2SO4 + 2 H2O
    Thus you are finding the Normality arising from the presence of Potassium Sulphate.
  2. Find the mass of Potassium Sulphate produced.
    2.45/98 = 0.025 moles of Sulphuric Acid are mixed with 2.805/56.1 = 0.05 moles of KOH.
    Thus 0.025 moles of Potassium Sulphate are produced.
    Mass of Sulphate produced = Moles x RMM = 0.025 x 174.2 g = 4.355 g
  3. Find the number of equivalents of Potassium Sulphate in 650 ml of solution.
    1 molecule of Potassium Sulphate can replace 2 atoms of Hydrogen.
    Thus the gram equivalent of Potassium Sulphate is RMM/2 ie 174.2/2 g = 87.1 g.
    Number of Equivalents =      Total Weight of Sulphate      =   4.355   = 0.05 Eq
    Sulphate Gram Equivalent          87.1
  4. 650 ml of solution contains 0.05 Equivalents. Thus:
    1000 ml of solution contains 1000/650 x 0.05 Eq = 0.076923 Eq = 0.07692 Eq to 5 dec pl
    The concentration is 0.07692 Eq/litre, giving a Normality of 0.07692.

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