Buffers that are usually used consist of either a weak acid and its conjugated base or a weak base and its conjugated acid.
Buffer Solution – Examples
- HCO3– /CO32- buffer
- H2PO4– / HPO42- buffer
- CH3COOH / CH3COO– Na+ buffer
Preparation of Buffer Solution
When we want to prepare buffer solution, we take,
- An acid which has the closest pKa value to the required pH and its conjugate base salt or,
- A base which has the closest 14-pKb value to the required pH and its conjugate acid salt.
The buffer ratio of the two components is found by the famous Henderson- Hassel Bach equation
pH = pKa + log B/A (B= base, A=acid)
Buffer capacity tells us how strong the buffer is in terms of withstanding any addition of base or acid. This depends on two things, the buffer ratio and the actual concentrations of the two components. A buffer is strong when both components are in equal concentrations. This happens only if pH = pKa or 14-PKb. Also, we know the buffer ratio could be obtained in multiple ways.
For example, if B/A = 2 then it could be obtained by,
B = 2 moldm-3 and A = 1 moldm-3
B = 0.2 moldm-3 and A = 0.1 moldm-3
B = 0.02 moldm-3 and A = 0.01 moldm-3 and so on.
However, it has been shown that higher the concentrations, higher the buffer capacity.
Buffer capacity is given by buffer index β.
β = ∆B/∆pH
∆B = amount of strong acid or strong base added (in molarity moldm-3) and ∆pH = the pH difference caused by the addition of a strong base or strong acid.
How to Calculate Buffer Capacity
STEP 1: Take 1 dm3 of the buffer of interest (1 Liter)
STEP 2: Measure the initial pH by using an accurately calibrated pH meter, pHx.
STEP 3: Add a known amount of strong acid / strong base and mix the solution well allowing equilibrating.
STEP 4: Measure the final pH of the mixture by an accurately calibrated pH meter, pHy.
STEP 5: CALCULATION
To illustrate I use acetic buffer, added amount of NaOH = 0.02 mol, pH x= 4.75 pHy= 5.20
Substituting to the equation,
β = ∆B/∆pH = 0.02 mol / (5.20-4.75)
NOTE: Always the buffer capacity is calculated for 1 dm3 of the buffer. If you use a different buffer volume, always calculate accordingly.
Say, the above pH difference was given by 500.0 cm3 of the buffer, then β = (0.044 mol/ 2) = 0.022 mol
Because the volume is ½ of 1dm3 the value obtained is multiplied by ½.
Leave a Reply