How to Calculate pH of a Buffer Solution





For a better understanding of how to calculate pH of a buffer solution, you have to know a buffer solution. What is a buffer solution? It is a solution that resists vicissitudes in solution pH when small amounts of either acid solution or basic solution are added to that solution (a buffer solution). Therefore, a buffer solution obeys Le Chatelier’s Principle. Le Chatelier’s Principle states that when a system at equilibrium is subjected to any change, the system readjusts itself to resist the effect (s) of the change. This results in the establishment of a new equilibrium. The change can either be pressure or temperature or concentration. 

A buffer solution can be an acid or a base. An acidic buffer solution resists the removal of hydrogen ions. On the other hand, an alkaline buffer resists the removal of hydroxide ions.

Up to this point, you can now calculate the pH of a buffer solution. However, you have to identify whether the buffer solution is an alkaline or an acidic buffer solution. 

Let us consider an example of a buffer solution. A buffer solution consists of 0.10 mol dm-3 of methanoic acid and 0.20 mol dm-3 sodium methanoate. Calculate the pH of this buffer solution. 

In a solution having a weak acid, equilibrium exists between undissociated acid and its ions. Hence, methanoic acid is represented in an equilibrium system as: 
H3COOH (aq) = H3COO- (aq) + H+ (aq)
Thus, the addition of methanoate ions will shift the equilibrium to the left. Though, the equilibrium has been shifted to the left, but the equilibrium will still occur. 

The equilibrium constant of the above equation can be written as:
Ka= [H3COO-][H+]/[H3COOH]

In calculations involving weak acid, it is assumed that the concentration of the species in the numerator of the equilibrium constant equation is the same. However, for equilibrium constant equation of a buffer solution, these species are not the same. Presence of methanoate ions further moves equilibrium to the left. But the methanoate ions originating from methanoic acid are insignificant than methanoate ions coming from sodium methanoate.    

Therefore, this leads to the assumption that methanoate concentration is equal to the concentration of sodium methanoate. 

Recommended books for further reading:
CLICK HERE to get a copy of Lehninger Principles of Chemistry By Nelson and Cox

In calculation involving weak acid, it is presumed that acid hardly ionized. Hence, at the equilibrium, the concentration of acid is equal to the concentration of acid used. So, this assumption aptly applies in this case (calculations of buffer solution). 

To calculate the pH of a buffer solution containing 0.10 dm-3 of methanoic acid and 0.20 mol dm-3 sodium methanoate, you have to know the Ka of methanoic acid. 

Ka of methanoic is 1.8 x 10-4 mol dm-3. Inserting the right values in the equilibrium constant equation gives the pH of the buffer solution by finding –log [H+]. 

Ka= [0.20 mol dm-3][H+]/0.10 mol dm-3
[H+]= 1.8 x 10-4 mol dm-3 x (0.1/0.20)
= 9 x 10-5 mol dm-3
pH=-log [= 9 x 10-5]
= 4.0 (in two significant figures)

The calculations of pH for alkaline buffer solution are the same as calculations involving acidic buffer solutions. 

Recommended books for further reading:
 CLICK HERE to get a copy of Organic Chemistry Concepts and Applications for Medicinal Chemistry 1st Edition By Rice
 
NB: The above pH can also be obtained using the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is derived from the equilibrium constant equation as follows: 

Let HB represent an acid while B- to denote conjugate base. Now, Ka (equilibrium constant equation) becomes:
Ka= [H+][B-]/[HB]

Calculation of hydrogen concentration [H+] gives:
[H+]= Ka[HB]/[B-]

By taking negative logarithms on both sides of the equation results in:
-log [H+]=-logKa - log[HB]/[B-]
Now, by substituting pH for log [H+] and pKa for –logKa gives:
pH=pKa – log[HB]/[B-]

Inverting part of the above equation (– log [HB]/ [B-]), which embroils changing its sign results in:
pH=pKa + log[B-]/[HB] ( Henderson-Hasselbalch equation)

Recommended books for further reading:
CLICK HERE to get a copy of Carbonic Acid Compounds and Hydrogen Ion Activities in Blood and Salt Solutions: A contribution to the Theory of the Equation of Lawrence J. Henderson and K.A Hasselbalch (Classical Reprint) By Warburg



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