Characterizing Solutions

PREVIOUS: Quantifying Compounds in Bulk

Having found a way to find number of molecules in solids in the previous post, its time to turn our attention toward solutions. Solutions are obtained by dissolving solute (such as salts, polymers, etc.) in a solvent. Such solutions are used in myriad of applications in research, industry as well as our homes. Consider the example of hypothetical lemonade we made in the last post wherein we had dissolved \(100 \, gm\) of sugar in \(1 \, liter\) of water. Since lemonade was prepared for four individuals, each individual would have got \(25 \, gm\) of sugar in their share along with \(0.25 \, liter\) or \(250 \, ml\) of water.

Now, it is not too hard to imagine that if we add some more sugar, say \(10 \, gm\), to the lemonade of one individual, she would find it sweeter compared to lemonades of her friends. This is because though the amount of sugar has increased from \(25 \, gm\) to \(35 \, gm\), the amount of water in her lemonade has not. To bring the sweetness of her lemonade on the same level as her friends', she would need to add water. But exactly how much water would she need to add?

I think it is clear that what determines the degree of sweetness of lemonade is the amount of sugar per unit volume of water. For example, prior to adding extra sugar to her lemonade, her lemonade had \(25 \, gm\) per \(250 \, ml\) water. Another way to put it would be: her lemonade had \(25 \, gm / 250 \, ml = 0.1 \, gm \, \mbox{of sugar}/ ml\). On adding \(10 \, gm\) sugar to this, the amount of sugar per \(ml\) of water increased to \(35 \, gm / 250 \, ml = 0.14 \, gm / ml\), resulting in a sweeter tasting lemonade.

Amount of sugar per \(ml\) of water can, thus, be used to describe the sweetness of lemonade. The higher the amount of sugar per \(ml\) of water, the sweeter-tasting the lemonade. We refer this quantity as \(concentration\) of the lemonade. To bring the sweetness back to the original level, she will need to reduce the concentration of lemonade from \(0.14 \, gm / ml\) back to \(0.1 \, gm / ml\). She can do this by adding more water: \[ \begin{array}{l} \frac{35 \, gm}{\mbox{volume of water}} &=& 0.1 \, gm/ml \\ \Rightarrow \mbox{Volume of water} &=& \frac{35 \, gm}{0.1 \, gm/ml} \\ &=& 350 \, ml \end{array} \] Therefore, she will need to add \(100 \, ml\) of water to her lemonade to make the total volume \(350 \, ml\), and the concentration \(35 \, gm / 350 \, ml = 0.1 \, gm / ml\). On doing so, her lemonade will taste just as sweet as it did prior to adding extra sugar.

Here, we described the concentration of lemonade in terms of 'mass of sugar per \(ml\) of water'. We can just as well describe the concentration in terms of the 'number of sugar molecules per \(ml\) of water' or 'number of moles of sugar per \(ml\) of water'. Here's how this can be done: In the previous post, we found that \[1 \, mole \, \mbox{of sugar has mass} \, 180 \, gm\]. This implies: \[\mbox{In} \, 25 \, gm \, \mbox{of sugar, there are} \, 25 \, gm/180 \, gm = 0.1389 \, moles \, \mbox{of sugar molecules}\]. Hence, the concentration of her lemonade is \[\begin{array}{l} \frac{0.1389 \, moles }{250 \, ml} = 0.0005556 \, mol/ml &=& 5.556 \times 10^{-4} \, mol/ml \\ &=& 5.556 \times 10^{-4} \times 6.023 \times 10^{23} \, molecules/ml \\ &=& 3.3464 \times 10^{20} \, molecules/ml \end{array}\] Thus there are \(3.3464 \times 10^{20}\) sugar molecules every \(ml\) of her lemonade. Adding \(10 \, gm\) extra sugar increase the total sugar molecules to , the concentration of her lemonade increases to \(35 \, gm/180 \, gm = 0.1944 \, moles\). Therefore, the concentration increases to: \[\begin{array}{l} \frac{0.1944 \, moles }{250 \, ml} = 0.0007778 \, mol/ml &=& 7.778 \times 10^{-4} \, mol/ml \\ &=& 7.778 \times 10^{-4} \times 6.023 \times 10^{23} \, molecules/ml \\ &=& 4.684 \times 10^{20} \, molecules/ml \end{array}\]

Preparing salt-water solution

Suppose you dissolve \(10 \, gm\) of common salt \(NaCl\) in \(150 \, ml\) of water. And now you would like to know exactly how many molecules of \(NaCl\) are there in every \(ml\) of water. The calculation is quite straightforward. First, figure out the molecular mass of \(NaCl\) in amu - one \(^{23}_{11}Na\) atom and one \(^{35.5}_{17}Cl\) atom together have \(58.5 \, amu\). This means \[ 58.5 \, gm \, \mbox{of} \, NaCl \, \mbox{has} \, 1 \, mole \, \mbox{molecules} \] \[ \Rightarrow 10 \, gm \, \mbox{has} \, 10 \, gm/58.5 \, gm = 0.1709 \, moles \] The concentration of salt solution, therefore, is \[ \frac{0.1709 \, moles}{150 \, ml} = 0.00114 \, mol/ml = \mbox{concentration of} \, NaCl \, \mbox{solution} \] Thus, there are 0.00114 moles of \(NaCl\) molecules per \(ml\) of water. Since \(NaCl\) dissociates into \(Na^{+}\) and \(Cl^{-}\) ions in water, we say there are 0.00114 moles of \(Na^{+}\) and \(Cl^{-}\) ions each in every \(ml\) of water.

Preparing \(NaOH\) solution

Sodium hydroxide (\(NaOH\)) solution is prepared by dissolving a specific amount of \(NaOH\) in a specific volume of water. Upon dissolution in water, \(NaOH\) dissociates into \(Na^{+}\) and \(OH^{-}\) ions, similar to the case of dissolution of\(NaCl\) in water. Thus, \(NaOH\) solution contains \(Na^{+}\) and \(OH^{-}\) ions. \(NaOH\) solution is as alkali regularly used in acid neutralisation and titration experiments. It can also be used to make insoluble compounds such as ferrous hydroxide (\(Fe(OH)_{2}\)) from \(FeCl_{2}\) or \(FeSO_{4}\) by ion-displacement method.

Suppose you dissolve \(20 \, gm\) of sodium hydroxide (\(NaOH\)) in \(150 \, ml\) of water to make \(NaOH\) solution of a specific concentration. To can use this solution in an experiment in a meaningful way, it is important to first find out its exact concentration, i.e., exactly how many molecules of \(NaOH\) are there in every \(ml\) of water. First, figure out the molecular mass of \(NaOH\) in amu - one \(^{23}_{11}Na\) atom, one \(^{16}_{8}O\) and one one \(^{1}_{1}H\) atom together have \(40 \, amu\). This means \[ 40 \, gm \, \mbox{of} \, NaOH \, \mbox{has} \, 1 \, mole \, \mbox{molecules} \] \[ \Rightarrow 20 \, gm \, \mbox{has} \, 20 \, gm/40 \, gm = 0.5 \, moles \] The concentration of salt solution, therefore, is \[ \frac{0.5 \, moles}{150 \, ml} = 0.00333 \, mol/ml = \mbox{concentration of} \, NaOH \, \mbox{solution}\] Thus, there are 0.00333 moles of \(NaOH\) molecules per \(ml\) of water. Since \(NaOH\) dissociates into \(Na^{+}\) and \(OH^{-}\) ions in water, we say there are 0.00333 moles of \(Na^{+}\) and \(OH^{-}\) ions each in every \(ml\) of water.

SUMMARY:
Solutions can be characterized using the quantity known as \(concentration\). \[\mbox{concentration of a solution} = \frac{\mbox{number of moles of solute}}{\mbox{volume of solution}}\].