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Solution Stoichiometry

The quantitative study of the relative amounts of reactants and products in chemical reactions is referred to as stoichiometry. Solution stoichiometry deals with relative quantities of reactants and products for chemical reactions occurring in solutions. 

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Solution Stoichiometry Definition

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Solution stoichiometry definition is given as study of quantities of reactants and products of chemical reactions taking place in solutions. These studies enable us to prepare solutions of desirable concentrations and to perform chemical reactions using accurate quantities of solutions. Solution stoichiometry is applied to predict the quantities of products to calculate the expected yields.

As we know that solutions are homogeneous mixtures and consist mainly of solvent and solute. Here, solvent is the major component being the solvent while solute constitutes the minor component. The most important property of solution is concentration which is defined as amount of solute dissolved in a given volume or weight of solution. We know that quantities of solute can be represented in more than on ways. Since quantities of solute and concentration are correlated, hence concentration of solution can also be expressed in different units. Some of the commonly used units for concentration of solution are weight of solute per unit volume of solution, quantity in mole per unit volume, weight percentage etc. Let’s look at some examples for expertise the conversion factors of solution concentration used in solution stoichiometry.

Example:
We know that at standard temperature and pressure, one mole of ideal gas occupies 22.4 L. Now, let’s calculate the concentration of this ideal gas in M. 
Concentration = $\frac{1 \ mol}{22.4 \ L}$ = 0.0446 M.

Solution Stoichiometry Dilution

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Solution stoichiometry dilution deals with diluting the concentration of solution by adding more solvent. Dilution refers to the process of lowering the concentration of the substances dissolved in a particular solution by adding water or solvent. The concept of dilution is practically applied to make solutions of any desired concentration even if the concentration of initial solution is much more than the desired one. Let’s take an example to understand the concept in better way. Let’s suppose that we have a glucose solution of 1.00 M. But for some purpose, we need to make a solution that is 0.50 M in glucose. It simply means that we need to reduce the concentration of that glucose solution in half. This desired concentration can be obtained by mixing equal volumes of the present 1.00 M glucose solution with distilled water. For example, if 1.0 liter of 1.0 M glucose solution is mixed with 1.0 liter of distilled water, the volume of solution is doubled to 2.0 liters. As we know that volume of a solution is inversely proportional to the solute concentration. This means that by increasing the volume of solution by double, the concentration of glucose in that solution is reduced to half i.e. 0.50 M.

Here, we should note that adding water to the solution does not disturb or change the amount of glucose. Neither adding water to the solution destroys glucose molecules nor does it affect them in any ways. This means that the number of moles of glucose in the original stock solution before the dilution is equal to the number of moles of glucose after the dilution by adding water. In the present example, the concentration of glucose in stock solution before dilution was 1.0 mole per liter * 1 liter, or 1.0 mole of glucose. The process of dilution by adding water altered the glucose concentration in solution as 0.5 moles per liter * 2 liters, or 1.0 mole of glucose. Solution dilution is expressed as following:
$C_{1} V_{1}$ = $C_{2} V_{2}$

Here, $C_1$ and $V_1$ represent the original concentration and volume of the solution before the dilution. The product of $C_1$ and $V_1$ represents the number of moles of glucose present before the dilution. Likewise, Here, $C_2$ and $V_2$ represent the concentration and volume of the solution after the dilution and their product represents the number of moles of glucose present after the dilution. Since the right hand side of equation is equal to the left hand side which means that the dilution does not change the moles of glucose or other solute.

How to do Solution Stoichiometry?

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Now, let’s learn how to do solution stoichiometry. As mentioned earlier, since quantities of solute can be expressed in multiple units and concentration of solution is dependent on the solute quantities; hence, we can deduce that solution concentration can also be expressed in multiple units. The two most common conventions to express the solution concentration are molarity and molality. Molarity is the molar concentration (M) of a solution. It is defined as the number of moles of solute (n) in one liter of solution. Since one liter of solution denotes volume of solution, hence molarity of solution is number of moles of solute per volume. It can be expressed as following equation:
M = $\frac{n}{V_{solution}}$.

Molarity is expressed in mol/L which is also abbreviated as M. Now. Let’s apply concept of molarity in solution stoichiometry problems.

If we want to calculate the number of moles of NaCl in 0.50L of a 1.00M solution of NaCl; using the formula for molarity:

= $\frac{n}{V_{solution}}$

Therefore, 0.50 L of solution $\times$ $\frac{1.00 \ mole}{1.00 \ L \ of \ solution}$ = 0.50 moles NaCl

Solution concentration can also be expressed as Molality. Molality is the molal concentration (m) of a solution. It is defined as the number of moles of solute (n) per kilogram of solvent. Therefore, molality uses the mass of the solvent. Molality equation can be expressed as following:
m = $\frac{n}{V_{solution}}$

Molality is expressed in mol/kg and is also abbreviated as “m”.

Example: To calculate the number of moles of NaCl dissolved in 0.50 kg of water (the solvent), if we need to make a 1.00m solution of NaCl, the concept of molality can be used.

As we know as,
m = $\frac{n}{m_{solvent}}$

Therefore, 0.50 kg of solvent $\times$ $\frac{1.00 \ mole}{1.00 \ kg \ of \ solvent}$ = 0.50 moles NaCl

Solution Stoichiometry Problems

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Let’s suppose that a lab experiment requires 6 L of a 0.75 M solution of HCl. The available stock solution is 20 M. How much stock solution is needed to be added to water to make the required solution?

Let’s apply the concept of dilution as mentioned above. Here, $M_1$ = 20 M, $M_2$ = 0.75 M, $V_2$ = 6 L and $V_1$ = $X$. Now, let’s put the values in the equation.

$M_{1} V_{1}$ = $M_{2} V_{2}$

20 M $\times$ $X$ = 0.75 M $\times$ 6 L

Or, X = 0.75 M $\times$ $\frac{6 \ L}{20 \ M}$ = 0.225 L
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