Liquid Solution : Methods of Expressing Concentration

 

LIQUID SOLUTION

A solution is a homogenous mixture of two or more substances, the composition of which may vary within limits. “A solution is a special kind of mixture in which substances are intermixed so intimately that they can not be observed as separate components”. The dispersed phase or the substance which is to be dissolved is called solute, while the dispersion medium in which the solute is dispersed to get a homogenous mixture is called the solvent. A solution is termed as binary, ternary and quartenary if it consists of two, three and four components respectively.

 

 Methods of expressing concentration of solution

Concentration of solution is the amount of solute dissolved in a known amount of the solvent or solution. The concentration of solution can be expressed in various ways as discussed below.

(1)     Percentage : It refers to the amount of the solute per 100 parts of the solution. It can also be called as parts per hundred (pph). It can be expressed by any of following four methods :

(i)      Weight to weight percent (% w/w)  \[=\frac{\text{Wt}\text{. of solute}}{\text{Wt}\text{. of solution}}\times 100  

e.g., 10% Na2CO3 solution w/w means 10g of Na2CO3 is dissolved in 100g of the solution. (It means 10g Na2CO3 is dissolved in 90g of H2O)

(ii)     Weight to volume percent (% w/v)  \[=\frac{\text{Wt}\text{. of solute}}{\text{Volume of solution}}\times 100  

e.g., 10% Na2CO3 (w/v) means 10g Na2CO3 is dissolved in 100 cc of solution.

(iii)    Volume to volume percent (% v/v)  \[=\frac{\text{Vol}\text{. of solute}}{\text{Vol}\text{. of solution}}\times 100  

e.g., 10% ethanol (v/v) means 10 cc of ethanol dissolved in 100 cc of solution.

(iv)    Volume to weight percent (% v/w)  \[=\frac{\text{Vol}\text{. of solute}}{\text{Wt}\text{. of solution}}\times 100 

e.g., 10% ethanol (v/w) means 10 cc  of ethanol dissolved in 100 g of solution.

 

(2)     Parts per million (ppm) and parts per billion (ppb) : When a solute is present in trace quantities, it is convenient to express the concentration in parts per million and parts per billion. It is the number of parts of solute per million (106) or per billion (109) parts of the solution.  It is independent of the temperature.

\[ppm=\frac{\text{mass of solute component}}{\text{Total mass of solution}}\times {{10}^{6}} ;

\[ppb=\frac{\text{mass of solute component}}{\text{Total mass of solution}}\times {{10}^{9}}

(3)     Strength : The strength of solution is defined as the amount of solute in grams present in one litre (or dm3) of the solution. It is expressed in g/litre or (g/dm3).

\[\text{Strength}=\frac{\text{Mass of solute in grams}}{\text{Volume of solution in litres}}

(4)     Normality (N) : It is defined as the number of gram equivalents (equivalent weight in grams) of a solute present per litre of the solution. Unit of normality is gram equivalents litre–1. Normality changes with temperature since it involves volume. When a solution is diluted x times, its normality also decreases by x times. Solutions in term of normality generally expressed as,

N = Normal solution; 5N = Penta normal, 10N = Deci normal; N/2 = semi normal

N/10 = Deci normal; N/5 = Penti normal

N/100 or 0.11 N = centinormal, N/100 or 0.001 = millinormal

Mathematically normality can be calculated by following formulas,

(i)     \[\text{Normality (}N\text{)}=\frac{\text{Number of g}.\text{eq}\text{.}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution }(l)}  \[=\frac{\text{Weight of solute in }g\text{.}}{\text{g}\text{. eq}\text{. weight of solute}\times \text{Volume of solution (}l\text{)}}     

(ii)     \[N=\frac{\text{Wt}\text{. of solute per litre of solution }}{\text{g}\text{eq}\text{. wt}\text{. of solute}}

(iii)     \[N=\frac{\text{Wt}\text{. of solute}}{g.\text{eq}\text{. wt}\text{. of solute}}\times \frac{1000}{\text{Vol}\text{. of solution in }ml}

(iv)     \[N=\frac{\text{Percent of solute }\times \text{ 10}}{\text{g}\text{eq}\text{. wt}\text{. of solute}}

(v)      \[N=\frac{\text{Strength in }g {{l}^{-1}}\text{ of solution}}{\text{g}\text{eq}\text{. wt}\text{. of solute}}

(vi)     \[N=\frac{Wt\times density\times 10}{Eq.\,\,Wt}

(vii)   If volume V1 and normality N1 is so changed that new normality and volume N2 and V2 then,

N1V1 = N2V2 (Normality equation)

(viii) When two solutions of the same solute are mixed then normality of mixture (N) is

\[N=\frac{{{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}}

(ix)    Vol. of water to be added i.e., (V2 – V1) to get a solution of normality N2 from V1  of normality N1

\[{{V}_{2}}-{{V}_{1}}=\left( \frac{{{N}_{1}}-{{N}_{2}}}{{{N}_{2}}} \right){{V}_{1}}

(x)     If Wg of an acid is completely neutralised by V ml of base of normality N

\[\frac{\text{Wt}\text{. of acid}}{\text{g}\text{eq}\text{. wt}\text{. of acid}}=\frac{VN}{1000}  ; Similarly,

\[\frac{\text{Wt}\text{. of base}}{\text{g}\text{eq}\text{. wt}\text{. of base}}=\frac{\text{Vol}\text{. of acid }\times N\text{of acid}}{1000}

(xi)    When Va ml of acid of normality Na is mixed with Vb ml of base of normality Nb

           (a)    If VaNa = VbNb (Solution neutral)

           (b)    If VaNa > VbNb (Solution is acidic)

           (c)     If VbNb > VaNa (Solution is basic)

(xii)   Normality of the acidic mixture  \[=\frac{{{V}_{a}}{{N}_{a}}+{{V}_{b}}{{N}_{b}}}{({{V}_{a}}+{{V}_{b}})}

(xiii) Normality of the basic mixture  \[=\frac{{{V}_{b}}{{N}_{b}}+{{V}_{a}}{{N}_{a}}}{({{V}_{a}}+{{V}_{b}})}

(xiv)  \[N=\frac{\text{No}\text{. of meq * of solute}}{\text{Vol}\text{. of solution in }ml}

(*1 equivalent = 1000 milliequivalent or meq.)

 

(4)   Molarity (M) : Molarity of a solution is the number of moles of the solute per litre of solution (or number of millimoles per ml. of solution). Unit of molarity is mol/litre or mol/dm3 For example, a molar 1M solution of sugar means a solution containing 1 mole of sugar (i.e., 342 g or 6.02 × 1023 molecules of it) per litre of the solution. Solutions in term of molarity generally expressed as,

1M = Molar solution, 2M = Molarity is two,

\[\frac{M}{2}  or 0.5 M = Semimolar solution,

 \[\frac{M}{10}  or 0.1 M = Decimolar solution,

\[\frac{M}{100} or 0.01 M = Centimolar solution

\[\frac{M}{1000} or 0.001 M = Millimolar solution

  • Molarity is most common way of representing the concentration of solution.
  • Molarity is depend on temperature as,  \[\text{Molarity}\propto \frac{1}{\text{Temperature}}
  • When a solution is diluted (x times), its molarity also decreases (by x times)

Mathematically molarity can be calculated by following formulas,

(i)     \[M=\frac{\text{No}\text{. of moles of solute }(n)}{\text{Vol}\text{. of solution in litres}}

(ii)     \[M=\frac{\text{Wt}\text{. of solute (in gm) per litre of solution}}{\text{Mol}\text{. wt}\text{. of solute}}

(iii)     \[M=\frac{\text{Wt}\text{. of solute (in gm)}}{\text{Mol}\text{. wt}\text{. of solute}}\times \frac{\text{1000}}{\text{Vol}\text{. of solution in }ml\text{.}}

(iv)     \[M=\frac{\text{No}\text{. of millimoles of solute}}{\text{Vol}\text{. of solution in }ml}

(v)     \[M=\frac{\text{Percent of solute }\times \text{10}}{\text{Mol}\text{. wt}\text{. of solute}}

(vi)     \[M=\frac{\text{Strength in }g{{l}^{-1}}\text{of solution}}{\text{Mol}\text{. wt}\text{. of solute}}

(vii)    \[M=\frac{\text{10}\times \text{Sp}\text{. gr}\text{. of the solution }\times \text{Wt}\text{. of the solute}}{\text{Mol}\text{. wt}\text{. of the solute }}

(viii)  If molarity and volume of solution are changed from M1V1 to M2, V2. Then,

M1V1 = M2V2 (Molarity equation)

(ix)    In balanced chemical equation, if n1 moles of reactant one react with n2 moles of reactant two. Then,  \[\frac{{{M}_{1}}{{V}_{1}}}{{{n}_{1}}}=\frac{{{M}_{2}}{{V}_{2}}}{{{n}_{2}}}

(x)     If two solutions of the same solute are mixed then molarity (M) of resulting solution.

\[M=\frac{{{M}_{1}}{{V}_{1}}+{{M}_{2}}{{V}_{2}}}{({{V}_{1}}+{{V}_{2}})}

(xi)    Volume of water added to get a solution of molarity M2 from V1 ml of molarity M1 is

\[{{V}_{2}}-{{V}_{1}}=\left( \frac{{{M}_{1}}-{{M}_{2}}}{{{M}_{2}}} \right){{V}_{1}}

Relation between molarity and normality

Normality of solution = molarity  \[\times \frac{\text{Molecular}\text{mass}}{\text{Equivalent mass}}

Normality × equivalent mass = molarity × molecular mass

For an acid, \[\frac{\text{Molecular mass}}{\text{Equivalent mass}} = basicity

So, Normality of acid = molarity × basicity.

For a base, = Acidity

So, Normality of base = Molarity × Acidity.

 

(5)     Molality (m) : It is the number of moles or gram molecules of the solute per 1000 g of the solvent. Unit of molality is . For example, a 0.2 molal (0.2ml) solution of glucose means a solution obtained by dissolving 0.2 mole of glucose in 1000gm of water. Molality (m) does not depend on temperature since it involves measurement of weight of liquids. Molal solutions are less concentrated than molar solution.

Mathematically molality can be calculated by following formulas,

(i)    \[m=\frac{\text{Number of moles of the solute}}{\text{Weight of the solvent in grams}}\times 1000=  \[\frac{\text{Strength per 1000 grams of solvent}}{\text{Molecular mass of solute}}

(ii)    \[m=\frac{\text{No}\text{. of }gm\text{ moles of solute}}{\text{Wt}\text{. of solvent in }kg}

(iii)   \[m=\frac{\text{Wt}\text{. of solute}}{\text{Mol}\text{. wt}\text{. of solute}}\times \frac{1000}{\text{Wt}\text{. of solvent in }g}

(iv)    \[m=\frac{\text{No}\text{. of millimoles of solute}}{\text{Wt}\text{. of solvent in }g}

(v)    \[m=\frac{\text{10}\times \text{solubility}}{\text{Mol}\text{. wt}\text{. of solute}}

(vi)    \[m=\frac{\text{1000}\times \text{wt}\text{. of solute (}x\text{)}}{\text{(100}-x\text{) }\times \text{ mol}\text{. wt}\text{. of solute}}

(vii)    \[m=\frac{\text{1000}\times \text{Molarity}}{\text{(1000}\times \text{sp}\text{. gravity)}-\text{(Molarity }\times \text{ Mol}\text{. wt}\text{. of solute)}}

Relation between molarity (M) and molality (m)

Molality (m) =  \[\frac{\text{Molarity}}{\text{Density}-\frac{\text{Molarity }\times \text{ molecular mass}}{\text{1000}}}

Molarity (M)  \[=\frac{\text{Molality }\times \text{ density}}{\text{1}+\frac{\text{Molality }\times \text{ molecular mass}}{1000}}

 

(6)     Formality (F) : Formality of a solution may be defined as the number of gram formula masses of the ionic solute dissolved per litre of the solution. It is represented by F. Commonly, the term formality is used to express the concentration of the ionic solids which do not exist as molecules but exist as network of ions. A solution containing one gram formula mass of solute per litre of the solution has formality equal to one and is called formal solution. It may be mentioned here that the formality of a solution changes with change in temperature.

Formality

(F) =  \[\frac{\text{Number of gram formula masses of solute}}{\text{Volume of solution in litres}} = \[\frac{\text{Mass of ionic solute (}g\text{)}}{\text{(}gm\text{. formula mass of solute) }\times \text{(Volume of solution (}l)\text{)}}

Thus,     \[F=\frac{{{W}_{B}}(g)}{GFM\times V(l)}\,\,\,\text{or}\,\,\,\frac{{{W}_{B}}(g)\times 1000}{GFM\times V(ml)}

 

(7)     Mole fraction (X) : Mole fraction may be defined as the ratio of number of moles of one component to the total number of moles of all the components (solvent and solute) present in the solution. It is denoted by the letter x. It may be noted that the mole fraction is independent of the temperature. Mole fraction is dimensionless. Let us suppose that a solution contains the components A and B and suppose that WAg of A and WBg of B are present in it.

Number of moles of A is given by, \[{{n}_{A}}=\frac{{{W}_{A}}}{{{M}_{A}}} and the number of moles of B is given by, \[{{n}_{B}}=\frac{{{W}_{B}}}{{{M}_{B}}}

Where MA and MB are molecular masses of A and B respectively.

Total number of moles of A and B = nA + nB

Mole fraction of A, \[{{X}_{A}}=\frac{{{n}_{A}}}{{{n}_{A}}+{{n}_{B}}} ; Mole fraction of B,  \[{{X}_{B}}=\frac{{{n}_{B}}}{{{n}_{A}}+{{n}_{B}}}

The sum of mole fractions of all the components in the solution is always one.

\[{{X}_{A}}+{{X}_{B}}=\frac{{{n}_{A}}}{{{n}_{A}}+{{n}_{B}}}+\frac{{{n}_{B}}}{{{n}_{A}}+{{n}_{B}}}=1                  

Thus, if we know the mole fraction of one component of a binary solution, the mole fraction of the other can be calculated.

Relation between molality of solution (m) and mole fraction of the solute (XA).

\[{{X}_{A}}=\frac{m}{55.5+m}                             

 

(8)     Mass fraction : Mass fraction of a component in a solution is the mass of that component divided by the total mass of the solution. For a solution containing WA gm of  and WB gm of B

\[\text{Mass fraction of }A=\frac{{{w}_{A}}}{{{w}_{A}}+{{w}_{B}}} ;

\[\text{Mass fraction of }B=\frac{{{w}_{B}}}{{{w}_{A}}+{{w}_{B}}}

Note :  It may be noted that molality, mole fraction, mass fraction etc. are preferred to molarity, normality, etc. because the former involve the weights of the solute and solvent where as later involve volumes of solutions. Temperature has no effect on weights but it has significant effect on volumes.

 

(9)     Demal unit (D) : The concentrations are also expressed in “Demal unit”. One demal unit represents one mole of solute present in one litre of solution at 0oC.