## General Chemistry

In the previous post, we talked about molarity which is the ratio of the moles of solute over the volume of the solution.

For example, if we dissolve  43.6 g K2SO4 in 1 L of water, the molarity of the salt is:

${\rm{n}}\;{\rm{(}}{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}\;{\rm{ = }}\;{\rm{43}}{\rm{.6}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{174}}{\rm{.3}}\;{\rm{g}}}}{\rm{ = }}\;{\rm{0}}{\rm{.250}}\;{\rm{mol}}$

${\rm{M}}\,{\rm{(}}{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{) = }}\;\frac{{\rm{n}}}{{\rm{V}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.250}}\;{\rm{mol}}}}{{{\rm{1}}\;{\rm{L}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.250}}\,M$

However, we know that when a strong electrolyte is dissolved in water, it dissociates into ions and the salt does not exist in the solution in its molecular form.

The concentration of the ions can be calculated from the concentration of the salt. For this, we need to identify how many of each ion appears in one molecule of the salt. So, for K2SO4, there are two K+ and one SO42- ion. You can also see this by writing the dissociation equation:

K2SO4(aq)  →  2K+(aq) + SO42-(aq)

Therefore, 0.250 M K2SO4 will produce two times more K+ ions and the same concentration of SO42- ions. The total concentration of all the ions will be 0.500 M K+ and 0.250 M SO42- = 0.750 M.

The concentration of an ion or the total ionic concentration can also be done using molar conversions.

For the concentration of K+ ions, we will have:

${\rm{M}}\;{\rm{(}}{{\rm{K}}^{\rm{ + }}}{\rm{)}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.250}}\;\cancel{{{\rm{mol}}\;{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}}}{{{\rm{1}}\;{\rm{L}}}}\;{\rm{ \times }}\,\frac{{{\rm{2}}\;{\rm{mol}}\;{{\rm{K}}^{\rm{ + }}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}}}\, = \;0.500\;M$

For the total concentration of all ions:

${\rm{M}}\;{\rm{(Ions)}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.250}}\;\cancel{{{\rm{mol}}\;{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}}}{{{\rm{1}}\;{\rm{L}}}}\;{\rm{ \times }}\,\frac{{{\rm{3}}\;{\rm{mol}}\;{\rm{ions}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}}}\, = \;0.750\;M$

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#### Practice

1.

What is the total concentration of ions in a solution prepared by dissolving 60.08 g of AlCl3 in enough water to make 600. mL of solution?

3.00 mol/L Al3+ and Cl

Solution

2.

How many potassium ions are present in 350. mL of 0.150 M K3PO4 solution? Consider K3PO4 as a strong electrolyte and ignore the reaction of phosphate ion with water. The molar mass of potassium phosphate is 212.3 g/mol.

9.48 × 1022 potassium ions

Solution

3.

Calculate the concentration of each ion and their total concentration in a solution prepared by dissolving 0.2500 mol of Ca(NO3)2 in 150.0 mL of water.

M (Ca2+)  = 1.667 mol/L

M (NO3)  =  3.334

Total concentration of all ions  =  5.0 M

Solution

First, let’s write down the dissociate equation of Ca(NO3)2:

Ca(NO3)2 → Ca2+ + 2NO3

Because each mole of Ca(NO3)2 produces one mol of Ca2+, their concentration is also going to be the same. To find the concentration of Ca(NO3)2, convert the volume from mL to L and use in the formula for molarity:

V (L) = 150.0 mL x 1 L/1000 mL = 0.1500 L

M (Ca(NO3)2) = n/V = 0.2500 mol / 0.1500 L = 1.667 mol/L

Therefore, the concentration of Ca2+ ions is also 1.667 mol/L.

On the other hand, each mole of Ca(NO3)produces 2 moles of NO3– ions. So, the concentration of NO3– ions = 2 x 1.667 mol/L = 3.334 mol/L.

The total concentration of all ions  = M (Ca2+) + M(NO3) = 1.667 M + 3.334 M = 5.0 M