-A buffer is an aqueous solution that has a highly stable pH. If you add acid or base to a buffered solution, its pH will not change significantly. Similarly, adding water to a buffer or allowing water to evaporate will not change the pH of a buffer. Examples of buffer problems:
Problem example 1: The first step involves calculating the proportion and amounts of the two ionic species in the buffer. The buffer contains a total of 10 litres x 0.045 M = 0.45 mole of phosphate (PO43-). pH = pKa2 + log [HPO42-]/[ H2PO4-]
7.5 = 7.2 + log [HPO42-]/[ H2PO4-] , 0.3 = log [HPO42-]/[ H2PO4-] Antilog of 0.3 = 2 = 2/1 ratio = [HPO42-]/[ H2PO4-] (NOTE: Always add 1 to the denominator to convert the number to a ratio) Therefore, mole of HPO42- needs = 2/3 x 0.45 mole = 0.30 mole Mole of H2PO4- needs = 1/3 x 0.45 mole = 0.15 mole Because we want to end up with an [HPO42-]/[ H2PO4-] ratio of 2/1, we want to convert ONLY 1/3 of the HPO42-to H2PO4- Mole of H2PO4- needs = mole of HCl needs = 0.15 mole Vol of 2.0 M of HCl = 0.15 mole/2.0 M = 0.75 litre (final answer) Mole of HPO42- needs = mole of HPO42- + mole of H2PO4- = 0.45 mole Vol of 1.0 M K2HPO4 = 0.45 mole/1.0 M = 0.45 litre (final answer). problem example 2: How many moles of sodium acetate and acetic acid are required to prepare I liter of a buffer, pH 5.0, which is 0. 1 M in total available acetate (dissociated and undissociated). Acetic acid has a pK of 4.74. Solution: From Buffer Equation ,
5.0 = 4.74 + log [salt]/ [acid]
or log [salt]/[acid] = 0.26.
The ratio of the concentration of sodium acetate to acetic acid is the antilog 0.26, or 1.82; there are 1.82 molecules of NaAc for each molecule of HAc. The mole fraction of salt is 1.82/2.82 and the mole fraction of acid is 1.00/2.82. We are told that the total molar concentration of acetate is 0. 1 M Therefore, of the 0. 1 mole, the salt must provide its fraction, 1.82 / 2.82 x 0. 1 mole =...