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Weak Acid Weak Base pH calculation solved example
Weak Acid Base pH calculation  Solved Example
In a previous post entitled “Weak Acids and Bases – Calculate the pH of a weak acid” a general equation was derived for a weak acid base pH calculation (mathematical approach). A fourstep method was also proposed for weak electrolyte pH calculation (calculate the pH for a weak acid  calculate the pH for a weak base, chemical approach) (Fig. I.1).
Solved examples using the above fourstep method were presented in the post “Solving Weak Acid and Weak Base pH problems”
Some extra solved examples on weak acid (base) chemistry are shown below. The “chemical approach” method is used for the solution.
Fig. I.1: Flowchart showing how to calculate the pH of a weak acid solution
Example I.1
Determine the pH of a 0.1 Μ weak acid HA solution (k_{a} = 10^{5}). Suppose that 1 ml of this solution is added to a beaker that contains 99 ml of H_{2}O to produce a new solution with total volume 100 ml. What is the pH value of the new solution?
DATA

[HA] = 0.1 M
k_{a} = 10^{5}
V_{H2O }= 99 ml

UNKNOWNS

a) pH = ? b) (pH)_{NEW} = ;

Step 1: Write the ionization equilibrium reaction. The pH of a weak acid of known concentration has to be calculated
HA is a weak acid as the k_{a} value shows. It dissociates partially in water according to equation (1):
Step 2: Write down the equilibrium constant expression and the equilibrium constant:
k_{a} = [Η^{+}] . [A^{}] / [HA] = 10^{5} (2)
Step 3: a) Let us suppose that x M of HA dissociate. Then the equilibrium concentrations of the species involved are as follows:
HA

Η^{+}

Α^{}


Initial

0.1 Μ

0 Μ

0 Μ

Change

x M

+x M

+x M

Final
(at equilibrium)

(0.1–x) M

x M

x M

b) The moles of HA n_{HA} in the new solution can be calculated as follows:
1000 ml solution HA contains 0.1 moles «pure» ΗΑ
1 ml solution HA contains y = ; moles «pure» ΗΑ y = 10^{4}moles «pure» ΗΑ.
The concentration of ΗΑ in the new solution can be calculated as follows:
100 ml HA contain 10^{4}moles «pure» ΗΑ
1000 ml HA contain y = ; moles «pure» ΗΑ y = 10^{3}moles «pure» ΗΑ.
Therefore, the initial concentration of ΗΑ in the new solution is equal to:
[HA] = 10^{3} M (6)
From (6) and similarly to (α) the [Η^{+}] concentration in the new solution can be calculated:
k_{a} = (x). (x) /(0.001x) = 10^{5 } and x^{2} / 0.001 = 10^{5 }_{ } x = [H^{+}] = 10^{4}M (6)
Step 4:
a) From (2) and the equilibrium concentrations from the above table:
(x). (x) /(0.1x) = 10^{5} and x^{2} / 0.1 = 10^{5} and x = 10^{3} M (3)
(0,1x) ≈ x since HA is a weak acid and dissociates slightly in water (as the k_{a }shows).
The above assumption is valid since [HA_{dissociated}]/ [HA_{initial}] * 100 = 1% < 5%
Therefore [Η^{+}] = [A^{}] = 10^{3} M (4) and pH = log([H^{+}]) = log([10^{3}]) = 3
b) pH = log([H^{+}]) = log([10^{4}]) = 4
The pH of the weak acid Ha is equal to 4 (concentration [HA] = 0.1 M)
Relevant Posts  Relevant Videos
References
 J.N. Butler “Ionic Equilibrium – Solubility and pH calculations”, Wiley – Interscience, 1998
 JL. Burgot “Ionic Equilibria in Analytical Chemistry”, Springer Science & Business Media, 2012
 D. Harvey, “Modern Analytical Chemistry”, McGrawHill Companies Inc., 2000
 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
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