Chemistry Net: 11/01/2015 - 12/01/2015

Cubic Equation Calculator for Weak Acid-Base Equilibria

Cubic equation calculator - Weak Acids and bases

 

Cubic Equation Calculator for Weak Acid-Base Equilibria

The mathematical (algebraic) exact method for solving weak acid or weak base equilibrium problems has traditionally been less popular than the alternative approximate method, probably because of the inconveniences related to solving cubic equations. However, modern mathematics software or even spreadsheets like Excel 2010 handle such equations with ease, making the algebraic method more attractive than in the past.

The mathematical approach to solving weak acid-base equilibria has been presented in a previous post entitled  “Weak Acids and Bases – Calculate the pH of a weak acid (a general equation)”.

There are two equilibria present: i) the dissociation of water and ii) the dissociation of the weak acid ka

[H+] [OH-] = kw        (1)

[H+] [A-] = ka [HA]        (2)

If the concentration of the acid in solution is C (M, moles/l), a mass balance on the anion A gives:

C = [HA] + [A-]            (3)

and a charge balance gives:

[H+] = [OH-] + [A-]             (4)

Let us assume that ka, kw  and C are known. Then there are four equations and four unknowns (shown in red in equations (1) to (4)).

A general expression for [H+ ] (or for the pH of a weak acid) in this case would be an expression in terms of the known ka, kw  and C. Therefore, the unknowns have to be eliminated starting from [HA] or [OH-] that are contained in the least number of equations. Let us eliminate first [OH-]. Solving equation (1) for [OH-] and substituting it in equation (4):

[H+] [OH-] = kw    and  [OH-] = kw / [H+]   (1’)

By substituting (1’) to (4) eliminates [OH-]:

[H+] = [OH-] + [A-] = kw / [H+] + [A-]           (5)

Next let us eliminate [HA] by solving equation (3) for [HA] and substituting in equation (2):

C = [HA] + [A-]   and [HA] = C - [A-]     (3’)

Substitute (3’) in equation (2) and eliminate [HA]:

[H+] [A-] = ka [HA]  = ka (C - [A-])     (6)

Now let us eliminate [A-] by solving equation (5) for [A-] and substitute in equation (6):

[H+] = kw  / [H+] + [A-]  and  [A-] = [H+] - kw / [H+]     (7)

Substituting (7) in equation (6):

[H+] [A-] =ka [HA]  = ka (C - [A-]) and

[H+] ([H+] - kw / [H+])  = ka (C -[H+] - kw / [H+]) )      (8)

Rearranging (8) we get a cubic equation in [H+]:

[H+]3 + ka[H+]2 – ( kw + kaC) [H+] - kwka= 0       (9)

Using the known values of C, ka, kw equation (9) can be solved for [H+] and the pH can be calculated. In our days, solution of equations like (9) has become easier by using on line cubic calculators such as the one shown below.

For acids use the weak acids calculator. Select the weak acid dissociation constant from the table and copy and paste it in the ka box. Type the initial acid concentration C in moles/liter (M) in the box named C (molar). The [H+] concentration and the pH of the solution are shown at the bottom.

Fig. I.1: The cubic equation calculator for weak acid-base equilibria. By inserting values for the initial acid concentration C(Molar)  and for the acid dissociation constant (values are shown in the table for the most common weak acids) the [H+] and pH values of the solution are calculated


For acids use the weak acids calculator shown below.


For bases use the weak bases calculator shown below. Follow the instructions given above.


 


Relevant Posts

Solving Weak Acid and Weak Base pH problems

Weak Acid Weak Base pH calculation solved example

Weak Acids and Bases - Calculate the pH of a weak acid

Polyprotic Acids / pH Calculation

Self-ionization of water - Autoionization of water - The ion product of water (kw)


References

  1. J.N. Butler  “Ionic Equilibrium – Solubility and pH calculations”, Wiley – Interscience, 1998
  2. J-L. Burgot “Ionic Equilibria in Analytical Chemistry”, Springer Science & Business Media, 2012
  3. D. Harvey,  “Modern Analytical Chemistry”, McGraw-Hill Companies Inc., 2000
  4. J.N. Spencer et al., “Chemistry structure and dynamics”, 5th Edition, John Wiley & Sons, Inc., 2012

Key Terms

weak acids and bases, calculate the pH of a weak acid, calculate the pH of a weak base, weak acid base chemistry, weak acid base pH calculator,chemistry app calculator,weak acid base ph app calculator ,weak acid base pH chemistry applet,

 

Balancing Oxidation - Reduction reactions by the Oxidation Number Method

Balancing Oxidation / Reduction Reactions

Balancing Oxidation - Reduction reactions by the Oxidation Number

In many important chemical reactions, electrons are transferred from atom to atom. Such reactions are called either oxidation – reduction reactions or redox reactions. Many important chemical reactions involve oxidation and reduction. In a previous post entitled “Oxidation – Reduction (Redox) Reactions – Balancing Redox Reactions” several examples of redox reactions were given and the concept of oxidation number was discussed. Solved examples for the determination of oxidation numbers to atoms in  chemical compounds were given in the post “Oxidation Numbers: Assigning oxidation numbers to atoms in a chemical compound – Examples”.

A number of different techniques have traditionally been taught in beginning chemistry students for balancing chemical reactions. These methods are the following:

  • Inspection: Most students of chemistry are taught to balance simple chemical reactions by “inspection”. Each student develops his or her own personal method based on practice. However, it becomes insufficient when relatively complex redox reactions have to be balanced.
  • Oxidation Number Method: This is another approach for balancing redox reactions. Oxidation numbers are assigned to atoms oxidized and reduced. By applying the general rule that the change in oxidation must be equal to the change in reduction, coefficients for these species are obtained, after which those of the remaining reactants and products.
  • Half-Reaction Method: The net reaction is divided into two half-reactions, one for oxidation and the other for reduction. These half reactions are balanced separately, then combined to form a balanced reaction.
  • Algebraic Method: The unbalanced chemical equation is used to define a system of linear equations, which can then be solved to yield the stoichiometric coefficients.

 The oxidation number method is going to be presented in this post. The oxidation number method uses the fact that the amount of oxidation must equal the amount of reduction in the total chemical reaction. The basic steps are the following:

Step 1:   Write the correct molecular formula for each reactant and product

Step 2: Identify the elements that change oxidation number during the reaction. Write the oxidation numbers of these atoms above the appropriate chemical symbols on both sides of the equation.

Step 3:  Calculate the increase or decrease in the oxidation number per atom and for the entire molecule / ion in which it occurs. If these are not equal then multiply by suitable number so that these become equal.

Step 4:  Balance the other metal ions that do not change oxidation number

Step 5:  Add H+ (in acidic solutions) or OH- ions (in alkaline solutions) to the reaction on the appropriate side so that the total ionic charges of reactants and products are equal.

Step 6:  Make the numbers of hydrogen atoms in the expression on the two sides equal by adding H2O molecules to the reactants or products. Similarily, balance the O atoms.

 

Let us apply the above in the following example:

 

Example I.1

Balance the following reaction:

ΗCl    +    K2Cr2O7    →     KCl  +  CrCl3   +   Cl2

 

Step 1: Write the correct molecular formula for each reactant and product.

The correct molecular formulas are given in this case

 

Step 2: Identify the elements that change oxidation number during the reaction. Write the oxidation numbers of these atoms above the appropriate chemical symbols on both sides of the equation.

unbalanced reaction between HCl and K2Cr2O7

 

The Cl ion is oxidized (the oxidation number increases from -1 to 0) while the Cr ion is reduced (the oxidation number decreases from +6 to +3).

 

Step 3: Calculate the increase or decrease in the oxidation number per atom and for the entire molecule / ion in which it occurs. If these are not equal then multiply by suitable number so that these become equal.

unbalanced reaction between HCl and K2Cr2O7 including oxidation number difference

 

HCl and Cl2 are multiplied by 3 (coefficient 3) while K2Cr2O7 and Cl2 by 1 (coefficient 1).

unbalanced reaction between HCl and K2Cr2O7 including oxidation number difference and coefficients

 

Step 4:  Balance the ions that do not change oxidation number

 The K atoms in our case (their oxidation number remains constant)

unbalanced reaction between HCl and K2Cr2O7 including oxidation number difference and coefficients and K coefficients

 

Step 5:  Add H+ (in acidic solutions) or OH- ions (in alkaline solutions) to the reaction on the appropriate side so that the total ionic charges of reactants and products are equal.

The reaction is in acidic solution (HCl). HCl must be added on the reactants side to balance the 14 Cl atoms on the product’s side. Therefore the coefficient of HCl is going to be 14.

unbalanced reaction between HCl and K2Cr2O7 including oxidation number difference and coefficients and K coefficients and Cl coefficients

 

Step 6:  Make the numbers of hydrogen atoms in the expression on the two sides equal by adding H2O molecules to the reactants or products. Similarily, balance the O atoms.

Since there are 14 H’s on the left-hand side we have to add 7H2O on the right-hand side to balance them. The reaction is now properly balanced.

balanced reaction between HCl and K2Cr2O7

 

 


Relevant Posts


References

  1. D. Harvey,  “Modern Analytical Chemistry”, McGraw-Hill Companies Inc., 2000
  2. R.D. Brown, “Introduction to Chemical Analysis”, McGraw-Hill Companies Inc., 1982
  3. S. S. Zumdahl, “Chemical Principles”, 6th Edition, Houghton Mifflin Company (2009)

Key Terms

oxidation, reduction, redox, oxidation number, balancing redox reactions,