Ionization of Water

We learned in my previous post the different weak electrolytes and strong electrolytes.  Strong electrolytes are 100% dissociated or ionized in water while weak electrolytes only few % are ionized in water.  Example is the salt in water solution, using electrical conductivity apparatus we can tell that its a strong electrolyte because it can produce very bright light while acetic acid on the other hand is a weak electrolyte, it only produce dim light in electrical conductivity apparatus.  Water on the other hand is an example of  weak electrolyte, lets see what happens in when water is combined with another water molecule:

Water molecule reacts with another water molecule forming hydronium ion and hydroxide ion.  The double arrow indicates that a reversible reaction occur, hydronium ion and hydroxide ion can also combine forming water molecules.  This is called autoprotolysis (or autoionization)  of water.  The bond breaking between water molecule and ions combining forming water molecules continues until the same rate happens, and this is called state of chemical equilibrium.

We can write simply the ionization of water molecules as follows:

The concentration of hydrogen ion in pure water is  mole per liter at . 

Since the ionization of water gives one H+ and OH-  , therefore the ionization of hyroxide ion is also  mole per liter at the same temperature.  The product of the concentration of the H+ and the OH- is called the equilibrium constant of water (symbol Kw).  Chemist use bracket [ ] to represents the concentration, therefore [H+] is the concentration of hydrogen ion and [OH-] is the concentration of hydroxide ion.  Kw of water can be calculated using the formula:

Calculating the Kw of water

Based from the calculation above Kw (equilibrium constant) of water is . Since Kw is constant whenever there is a change in the concentration of hydrogen ion the concentration of hydroxide ion is also changes or vise versa.

Example:

Calculate the hydroxide ion concentration [OH-] given the hydrogen ion concentration [H+]:

Solution:

  TRY THIS:

Balancing Oxidation-Reduction Reaction

There are ways in which we can balance chemical equation, simple equations can be balanced using inspection method .   Simple oxidation-reduction reaction can also be balanced by inspection method but complicated oxidation-reduction reaction cannot be balanced by this method.  In this post I will be discussing two methods of balancing Redox reaction, the oxidation number method and ion-electron method.

Balancing Redox Reaction using Oxidation Number Method

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In balancing redox reaction using oxidation method, there are guidelines to follow:
1.  Examine the reactants and products to determine if there is a change in the oxidation number by calculating the oxidation number or by inspection.  Use the technique used in my previous blog about how to determine the oxidation number.

a.  Write the oxidation number of the element above its symbol.  See to it that all elements in the reactant side and in the product side have their oxidation number.

b.  Diagram the number of electron lost by the oxidized element and the number of electron gained by the reduced element.

2.  Balance each element using a coefficient and remember that the electron lost and gain are equal.

a. Place a coefficient before the formula of the oxidized substance that corresponds to the number of electrons gained by the reduced substance.

b.  Place a coefficient before the formula of the reduced substance that corresponds to the number of electrons lost by the oxidized substance.

c.  Balance the remaining elements by inspection.

3.  After balancing the equation, check if the coefficient of the reactants and products are correct.

a. Place a check above the  symbol of the element to indicate that the coefficient are correct.  Meaning the number of atoms in the reactant side is equal to the number of atoms in the product side.

b.  For ionic equation, make sure also that the charges both in the reactant side and product side are equal.


Let us have an example:

1. Balance using oxidation number method the reaction between  Copper and Silver nitrate forming Copper (II) nitrate and solid silver, as shown in the unbalanced equation below:

Step 1.  Write the oxidation number of each element:

Step 2.  Diagram the number of electrons lost and gained.

Step 3.  Place a coefficient before the formula of the oxidized substance that corresponds to the number of electrons gained by the reduced substance.

Based from the figure above Cu loses 2e-, therefore to balance the electrons gain coefficient of two is placed in front of AgNO3  and Ag. Since silver gains only 1 e- therefore only 1 is the coefficient of Cu

Step 4.  Place a check above the  symbol of the element to indicate that the coefficient are correct.  Meaning the number of atoms in the reactant side is equal to the number of atoms in the product side.

Since number of atoms are balanced both in the reactant and product side, therefore the equation is already balanced.

Balancing Equation Using Ion-Electron Method in Acidic and Basic Solution

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Ion-electron method is also called half reaction method, used to balance ionic redox reactions in acidic and basic solution. A half reaction is  a part of redox reaction that shows the element that undergoes oxidation and reduction process. The ion-electron method balances the oxidation half-reaction with the reduction half-reaction separately.

 There are several steps to be followed in balancing ionic redox equation by ion-electron method.

1.  Write the half-reaction both the oxidation and reduction. Write the reactant and the product of the element that undergoes oxidation, Also write the reactant and product of the element that undergoes reduction.

2.  Balance the element in each  half-reaction by writing coefficient.

     a.  Balance all elements except oxygen and hydrogen.

     b.  Balance oxygen using H2O.

     c.  Balance hydrogen using H+.

Please take note:  For reaction in basic solution, add  OH- is to neutralize H+.

                              For example 2OH- neutralizes 2H+ to form 2H2O molecules.

     d,  Balance the charge by using e-.

3.  Multiply each half-reaction by a whole number to balance the number of electrons lost in oxidation process and the number of electrons gained in reduction process.

4.  Add the two half-reactions together and cancel similar ions, including electrons on each side of the equation.

5.  After balancing, check the number of atoms of all the elements if balanced.  Also calculate the net charged if balance both in the reactant side and in the product side.

Let us have an example:

Example 1.  Write a balanced equation on the reaction below in acidic solution:

Solution:

Step 1.   Write the half-reaction both the oxidation and reduction. 

Step 2.  Balance all the elements and the charge.

Step 3.  Multiply the two half reaction with a whole number to balance the number of e-.  Since the first half-reaction has only one e- and the second half reaction has 5e-  then the first half-reaction must be multiplied by 5.

Step 4.  Add the two half-reactions together and cancel 5e- both in the reactant side and product side,

Step 5.  Verify the equation if balance by checking the number of atoms and charges.

               

   

Example 2.  Write a balance equation of the equation below in basic solution.

Solution:

Step 1.  Write the half-reaction the oxidation and reduction.

Step 2.   Balance all the elements, balance the O by adding H2O, and to balance H add H+.  Balance charges by adding e-. Since this is for basic solution neutralize H+  with OH-.

In basic solution H+ should be neutralize with the same number of OH-.

Step 3.  Balance the number of e- by multiplying whole number to the half reaction. Since the first half reaction lost 2e- and the second half-reaction gained 3e-,  to balance the lost and gained of electrons multiply the first half-reaction with 3 to make it 6e- and the second half-reaction 2 to balance the e-.

Step 4.  Add the two half-reactions, and cancel all similar ions including e-.

Step 5.  Check if the number of atoms and charges are equal.

TRY THIS:

Balance the following:

Oxidation Number

Oxidation number is also called as oxidation states, it refers to the electrons lost or gain during chemical reaction.  To determine whether an atom undergoes oxidation or reduction in a chemical reaction, oxidation number is determined before and after reaction. Let us see for example the reaction of H2 and Cl2 in the formation of HCl:

In the above equation, in the reactant side H2 has an oxidation number of 0, Cl2 has also oxidation number 0 while in the product side H has +1 oxidation number and Cl has -1 oxidation number.
H showed an increase of oxidation number therefore it is the one that undergoes oxidation or loses an electron while Cl showed a decrease of oxidation number therefore it undergoes reduction or gains an electron.

Now in order to identify if an atom undergoes oxidation or reduction, we need to assign oxidation numbers to the different atoms in the reactant side and in the product side. There are rules to follow in assigning oxidation numbers.  Below are the rules:

1.  In free elements, each atom has an oxidation number equal to zero.  Example Na, K, Mg, Cu,  etc. Diatomic molecules also have zero oxidation number like O2, Cl2, F2, N2, Cl2, Br2, H2,  including other molecules like F4, S8.

2.  For monoatomic ions there oxidation number is equal to the charge on their atoms.  You can use periodic table in determining the oxidation  number or the different list of monoatomic ions in my previous blog.  Example elements in Group 1A ions  have +1 oxidation number, group 2A ions have +2 oxidation number.

3.  Oxidation number of oxygen is always -2 but in peroxide compounds oxygen has a charge of -1. Example in H2O (water), H has a charge of +1 and O has a charge of -2 and H2O2 (hydrogen peroxide), H has +1 charge and O as a charge of -1.

There are two atoms of oxygen having -1 charge, that is why peroxide has -2 oxidation number.

4,  Oxidation number of hydrogen is always +1 when paired with other nonmetals or polyatomic anions except for hydride. Example in HCl, hydrogen has an oxidation number of +1.  Hydride has -1 oxidation number, example CaH2, hydrogen here has an oxidation number of -1.

5. Flourine has -1 oxidation number in all compounds.  Other atoms in the halogen group ( Cl, Br, I) also has -1 oxidation number except when they are bonded with oxygen forming polyatomic ions, their positive oxidation numbers are used.  Example :

6.  In a neutral molecule, the total charge of the atom must be equal to zero.  In polyatomic ions the total charge is equal to the charge of the ion.

 Example of neutral molecule is MgCl2,  

      Mg has a charge of +2 and Cl has a charge of -1 x 2 = -2.  Therefore +2 - 2  = 0 or neutral.   

Example of polyatomic ion,

What is the charge of Sulfur?

     S  (-2x4)  =  -2

     S  (-8)  =  -2

     S  =  -2  +  8

     S  =  +6

Therefore the oxidation number of S in  is + 6.

TRY THIS:

Identify the oxidation number of each element of the following compounds and ions:

CLICK HERE FOR THE ANSWER:

Oxidation-Reduction Reactions

Oxidation-Reduction reaction is also called redox reaction, which involves transfer of electrons between two species.  The oxidation number of atoms or ions change by either gaining or losing electron. Oxidation-reduction are so vital that it occurs in the different processes from our environment, plants and animals. Example of which is  the reaction of Magnesium and oxygen

2Mg(s)   +   O₂(g)    à    2MgO(s)

The example above involves the reaction of magnesium atom with the oxygen atom producing magnesium oxide. Magnesium oxide is made up of  Mg⁺² ion and O^-2 ion.  We have to analyze what happens to magnesium and oxygen in the reaction by trying to identify if they undergo oxidation or reduction.

             2Mg    à     2 Mg⁺²     +     4e^-

O₂    +    4e^‑    à     2O^-2

The above equations are the two half-reactions that occur in the reaction of magnesium and oxygen. Half-reaction shows the electrons involved in a redox reaction..  Two magnesium atoms gives up 4 electrons to oxygen atoms and oxygen atoms gain 4 electrons from magnesium atom. The sum of the two half-reactions give the overall reaction which is

2Mg   +   O₂   +  4e^-     à     2Mg⁺²     +     2O^-2     +    4e^-     

If we cancel electrons on both sides of the equations, it will look like this:

2Mg    +     O₂      à     2Mg⁺²     +     2O^-2

Checking the equation if balance, Mg has 2 atoms, O has also 2 atoms and both charges are equal to zero in the reactant side and in the product side.

Understanding about the two half-reactions, oxidation reaction, refers to the half-reaction that involves the loss of electron while a reduction reaction refers to the half-reaction that involves the gain of electrons.  In the above reaction, Mg is oxidized and O is reduced.  Mg is called the reducing agent, since it is the one responsible for the O to be reduced.  On the other hand O is the oxidizing agent since it is the one responsible for the Mg to be oxidized.  Therefore, reducing agent is the substance that donates electrons and is oxidized while oxidizing agent is the substance that accepts electrons and is reduced.

TRY THIS PROBLEM:

Write the half-reaction and identify the reducing and oxidizing agent of the following redox reaction:

1.       4Fe  +   3O₂     à      2Fe₂O₃

2.        Cl₂    +    2NaBr     à    2NaCl   +   Br₂ 

1.    3.        Si   +   2F₂     à    SiF₄

1.    4.        H₂   +   Cl₂    à    2HCl

Acid-Base Reaction

In my previous post, I discussed the different types of chemical reactions, the combination reaction, decomposition, single replacement and double replacement reaction. Acids and bases undergo reaction.  Acid-Base Neutralization is a kind of double replacement reaction producing salt and water.  Another reaction between acid and base  resulting to the production of gas.

Acid-Base Neutralization

Acid-Base Neutralization reaction is an acid and base reaction producing salt and water.   It is called neutralization since the acidity and basicity of acids and bases are neutralized. See the the format below:

When acid and base react it produce salt and water. It is called neutralization since the H+ ion concentration is being neutralized with OH- ion making it neutral.  The acidity and basicity of the two reactants are canceled out.

For example, when a solution of HCl is mixed with NaOH, neutralization occur producing water and salt.

Since both HCl and NaOH are strong electrolytes, both are completely ionized in water.  The ionic equation is shown below:

Therefore, the reaction can be represented by the net ionic equation as shown below:

Na+ and Cl-  are spectator ions.

Lets consider a reaction between a weak acid and a strong base,

The ionic equation of this reaction is:

HCN being a weak base will not be ionized in water.  And the net ionic equation of this reaction :

Other reactions of acids and bases are the following:

Acid-Base Reactions Leading to Gas Formation

Some salts that are basic when it reacts with an acid produced gas.  These salts are the following:

Examples of reaction between this salts and acids are shown below: