How many mL of a 0.10 M NaOH solution are needed to neutralize 15 mL of a 0.20 H_{3}PO_{4}?
Although there are several valid approaches to answering this question ( see other answers), practicing chemists have been using conveniently the following steps:
1. Conversion of all concentrations to normal concentrations ( N , number of gram equivalents/L ) (*)
2. Using the following formulae : Vx Nx = V N (1)
From (1)
Vx = V N/Nx (1a)
and Nx = N V/Vx (1b)
where V ,Vx and N,Nx are the volumes and the normal concentrations of a the known and unknown solutions respectively.
1b has general application in volumetric analysis (e.g. titrations) and 1a is useful in our case and in all preparative calculations for all chemical reactions.
step1. Conversion to normalities: 0.1 M NaOH = 0.1 N NaOH (*)
0.2 M H3PO4 = 0.6 N H3PO4 (*)
Step2. Substituting in 1a above: Vx =0.6/0.1 X 15 =90 ml. Of course if remembering or using formulae is not your way to work, a free thinking alternative is available. In this case also, conversion to normal concentration is indispensable for direct comparisons.
Here we have 15 ml of a solution that is 0.6N this means we have 15X0.6 =9 milliequivalents (mEq) of acid . Hence we will need 9 mEq of a base for a complete neutralization ,and we must get them from a 0.1N NAOH solution. Since 1 ml of this base solution contains 0.1 mEq, 9/0.1 =90 ml are needed.
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1. Conversion of all concentrations to normal concentrations ( N , number of gram equivalents/L ) (*)
2. Using the following formulae : Vx Nx = V N (1)
From (1)
Vx = V N/Nx (1a)
and Nx = N V/Vx (1b)
where V ,Vx and N,Nx are the volumes and the normal concentrations of a the known and unknown solutions respectively.
1b has general application in volumetric analysis (e.g. titrations) and 1a is useful in our case and in all preparative calculations for all chemical reactions.
step1. Conversion to normalities: 0.1 M NaOH = 0.1 N NaOH (*)
0.2 M H3PO4 = 0.6 N H3PO4 (*)
Step2. Substituting in 1a above: Vx =0.6/0.1 X 15 =90 ml. Of course if remembering or using formulae is not your way to work, a free thinking alternative is available. In this case also, conversion to normal concentration is indispensable for direct comparisons.
Here we have 15 ml of a solution that is 0.6N this means we have 15X0.6 =9 milliequivalents (mEq) of acid . Hence we will need 9 mEq of a base for a complete neutralization ,and we must get them from a 0.1N NAOH solution. Since 1 ml of this base solution contains 0.1 mEq, 9/0.1 =90 ml are needed.
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(*)The gram equivalent(E) of a molecule can be calculated as the ratio of the gram mole (M) and the valence (V) of the compound. E= M/V (2). The valence of a reactant in a neutralization reaction can be calculated as the product of the valence of the
group involved in the reaction and the number of groups present in the molecule.E.g. for the acid H2SO4, the group involved in acid-base reactions is the proton H+ that is monovalent. Since there are 2 of them, the valence of this acid is 1X2=2. Sulfuric
acid is a divalent acid. Hence E=M/2. And for H3PO4 , E=M/3. Phosphoric acid is a trivalent acid. Of course if the neutralization is only partial the valence is equally reduced.in the case of H3PO4 if only one H is neutralized as in the reaction H3PO4 + NaOH=
NaH2PO4 +H2O the valence of the acid is 1 because the number of groups (H+) participating is only one.