Ion Concentrations

I. Dissociation

                A. Ionic compounds are made up of 2 parts

                                a. Cation: positively charged particle

                                b. Anion: negatively charged particle

                B. When ionic compounds are dissolved in water the cation and anion separate from each other. This process is called dissociation

                C. When writing dissociation equation the atoms and charges must balance

                                a. The dissociation of sodium chloride is:

                                NaCl ---> Na⁺ + Cl^-

Ex. Complete the following dissociations

1. Fe(OH)₂ ---> Fe²⁺ + 2OH^-

2. Na₃PO₄ ---> 3Na⁺ + PO₄^3-

3. Fe₃(PO₄)₂ ---> 3Fe²⁺ + 2PO₄^3-

                        D. If the volume does not change then the concentration of individual ions depends on the balanced coefficients in the dissociation equation

Ex. Determine the [Na⁺] and [PO₄^3-] in a 1.5M solution on Na₃PO₄

Na₃PO₄ -->  3Na⁺ + PO₄^3-

1.5M x (1/1) = 1.5M [PO₄^3-]

1.5M x (3/1) = 4.5M [Na⁺]

Ex. Determine both ion concentrations when a 2.5M solution of Lithium sulphate dissociates.

Li₂SO₄ ---> 2Li⁺ + SO₄^2-

2.5M x (2/1) = 5.0M [Li^-]

2.5M x (1/1) = 2.5M [SO₄^2-]

Dilutions

I. Diluting Solutions

                A. When 2 solutions are mixed the concentration chenges

                B.  Dilution: process of decreasing the concentration of a solution by adding solvent (usually water)

                                a. The amount of solute doesn't change, only the volume changes

                                                i. v₁=v₂

                C. Because concentration is [mol/L] we can write:

                                a. C= n/v              or           n=C/v   so C₁v₁=C₂v₂

Ex. Determine the concentration when 100mL of 10.0M HCL is diluted to a final volume of 400mL

C₁= 0.1M

V₁=100mL

C₂=  ?

V₂= 400mL

C₁v₂=C₂v₂

C₁v₁ / v₂= C₂

(0.1M)(100mL) / (400mL) = 0.025 mol/ mL

Ex. How much water must be added to 10.0mL of 10.0M Na₂SO₄ to give a solution with a concentration of 0.50M?

C₁=10.0M

V₁= 10.0mL

C₂=0.50M

V₂=?

C₁V₁=C₂V₂

C₁V₁ / C₂= V₂

(10.0M)(10.0mL) / (0.5M) = 200mL

ΔV = V_f - V_i

ΔV = 200 - 10

ΔV = 190 mL of water needs to be added

Ex. How much water must be evaporated from 2.00L of 0.250M KCl solution for the final concentration to be 2.75M?

C₁=0.250M

V₁= 2.00L

C₂=2.75M

V₂=?

C₁V₁=C₂V₂

C₁V₁ / C₂= V₂

(0.250M)(2.00L) / (2.75M) = 0.182L

ΔV = V_f - V_i

ΔV = 0.182 - 2.00

ΔV = 0.182 - 200

ΔV= -1.82

Ex. [100 mL of 0.250M]_a sodium nitrate is mixed with [200mL of 0.100M]_b sodium nitrate. Find the concentration.

-Find the mol and volume of each solution separately

Solution a.  (0.100L) x (0.250mol / 1L) = 0.025 mol

Solution b. (0.200L) x (0.100mol / 1L) = 0.0200 mol

-Add together

mol_tot = 0.025mol + 0.0200mol = 0.0450mol

volume_tot= 100mL + 200mL = 300mL

(mol_tot) / (Volume_tot) = Molarity_tot

(0.0450mol) / (0.300L) = 0.150M

Ex. What mass of silver nitrate is needed to make 75.0mL of a 0.130M solution?

(0.130mol / 1L) x (0.075L) x (169.9g / 1mol) = 1.66g

Titrations

I. A titration is an experimental technique used to determine the concentration of an unknown solution

II. Terms & Equipment

                A. Buret - contains the known solution. Used to measure how much was added

                B.  Stopcock - valve used to control the flow of solution from the buret

                C. Pipet - used to accurately measure the volume of the unknown solution

                D. Erlenmeyer flask - container for unknown solution

                E. Indicator - Used to identify the end point of the titration

                F. Stock solution - known solution