Computing pH and pOH for Weak Acids and Bases
Calculating pH and pOH for Weak Acids and Weak Bases.
* Weak Acids and Weak Bases.
-unlike of those in strong acids and strong bases, weak acids and weak bases do not completely dissociate (dissociate only slightly in aqueous solution or in water) at equilibrium. The majority of the molecules remain undissociated.
*WEAK ACIDS
-acid ionization constant (Ka) is the equilibrium constant for an acid. At a given temperature, the strength of acid is measured quantitavely. You must first be familiar with equilibrium constant expressions and how to write them in a reaction. Then, by making an ICE (initial, change and at equilibrium) table, you can find unknown concentration values that can be get into this equilibrium expression.
EXAMPLES:
a) Calculate the pH of a 0.5 M HF (hydrofluoric acid) solution at 25 degree celsius with an ionization constant for HF is 7.1 x 10-4.
1. First, write the balanced equation for the given problem.
*the double-like arrow represents a reversible reaction which the product can be converted back to the original reactants under conditions and often means partially dissociation.
2. Using the data given above, make an ICE table, considering the ionization of hydrofluoric acid in water into fluorine ion and hydrogen ion. You can ignore the concentration of water (a pure liquid) in our calculation because only solutes and gases are used into the equilibrium expression.
* For every hydrofluoric acid molecule that dissociates, one fluorine ion and one hydrogen ion is produced which can be represented by subtracting "x" from the original hydrofluoric acid concentration, and adding "x" to the original concentrations of the dissociated ions.
3. Because Ka is given, you can use either the quadratic equation or the formula:
Then, substitute the values.
Do the operation. You can either use any of the formula; either the quadratic or the equilibrium constant for acids. After cross multiplying, you’ll come up with:
After getting the values for a, b, and c, substitute this value for the quadratic equation.
You’ll come up with the answer of:
To make sure that the value for x is valid, use the formula for the Ionization % = [H+] / [HA] x 100% : where HA is the main acid being dissociated. The ionization percent must not exceed 5% or it will be invalid.
You can now finally calculate the concentrations of H+, F- and HF at equilibrium..
You can now identify and compute for the pH of the solution using: pH = -log (H+)
[H+]= 0.02
pH = - log (0.003)
pH = -(-2.52)
pH = 2.52 WEAK ACID
* WEAK BASES:
-base ionization constant (Kb) is the equilibrium constant for a base.
EXAMPLE:
2. Determine and calculate the pH and pOH of a 7.0 x 10-3 M of NH3 (ammonia).
*Kb = 1.8 x 10-5
a) Write the balance equation and make an ICE diagram/table for the initial concentrations, change and at equilibrium.
b) Use the formula: Kb = [product] / [reactant] and get the values for a, b, and c. Use the quadratic equation and you’ll get an answer of:
c) To determine whether it is valid or not, use the formula for the ionization %: Kb = [OH-] / [HB] x 100%.
d) Substitute the acquired answer and get the concentrations at equilibrium.
[OH-] = 3.46 x 10-4
[NH4] = 3.46 x 10-4
[NH3] = 7.0 x 10-3 – 3.46 x 10-4
= 6. 65 x 10-3
e) Get and calculate the ph and also pOh:
[OH-] = 3.46 x 10-4
*pOH = -log (3.46 x 10-4)
pOH = - (-3.46) = 3.46
*pH + pOH = 14
pH = 14 – pOH
= 14 – 3.46 = 10. 54 WEAK BASE
Exercises:
1. Calculate the pH of a 0.036 M solution of Nitrous Acid (HNO2) with Kb of 8.3 x 10-5.
2. Calculate the pH of a 0.20 M aqueous solution of pyridine, C5H5N. The Kb for C5H5N is 1.8 x 10-9
Sources:
http://www.kentchemistry.com/links/AcidsBases/pHWeakBases.htm
Written by: Sydney Santiago
