Using your knowledge of equilibrium, pH, Ka and Kb values, choose a buffer pair
ID: 509799 • Letter: U
Question
Using your knowledge of equilibrium, pH, Ka and Kb values, choose a buffer pair from the acids, bases and salts provided in Table 2 (see graph below) to prepare two buffer solutions. (one of pH 9.0 and one of pH 7.0). Determine the correct molar ratio of acid or base to salt needed for the selected pH. Then determine the mass (for solids) or volume (for liquids) of the chemicals needed to make 100 mL of the buffer with a Buffering Capacity so that 50 mL of the buffer is capable of neutralizing at least 2 mL of either 1 M HCl or 1M NaOH without letting the pH change by more than 1 pH unit. See your text Sample Exercise 17.5, pages 733&734. You may also need to look up the solubility of each chemical. See the MSDS forms for this information.
Table 2. Available Acid-Base pairs for Preparation of buffer solutions in this experiment pKa PKD Formula Molecular Chemical Type Weight 60.05 1.8x 10-5 4.75 Acetic Weak acid acid Sodium Conjugate 82.03 acetate, base of anhydrous above acid Weak acid NaH2PO4 H20 137.99 6.2 x 10 7.21 Sodium phosphate monobasic Sodium Conjugate Na2HPO4 7H20 268.07 4.7 x 10 12.32 phosphate, base of above acid dibasic Citric acid Weak acid c6Hao, .H20 210.14 7.5 x 10 4 3.128 monohydrate Sodium citrate Conjugate Na3C6H507 .2H20 294.11 dihydrate base of above acid 17.3 5.9 x 10 10 9.24 1.7 x 10-5 4.76 NH40H Ammonia 2.0 M Weak base solution 53.5 Ammonium Conjugate NH4CI acid of aboveExplanation / Answer
Buffers are made up of an acid-base conjugate pair. Choose the acid-base pair where the pKa corresponds to the pH. From the given table, ammonia and ammonium chloride can be used to make pH 9 buffer and sodium phosphate monobasic and sodium phosphate dibasic can be used to prepare a buffer of pH 7.
Use Henderson-Hasselbach equation to find out the molar ratios of acid and base.
For pH 9,
pH = pKa + log ([A-] / [HA])
9 = 9.24 + log ([A-] / [HA])
-0.24 = log ([A-] / [HA])
([A-] / [HA] = 0.575
The ratio of ammonia to ammonium chloride should be 0.575.
For pH 7,
pH = pKa + log ([A-] / [HA])
7 = 7.21 + log ([A-] / [HA])
-0.21 = log ([A-] / [HA])
([A-] / [HA] = 0.617
The ratio of sodium phosphate dibasic to sodium phosphate monobasic should be 0.617.
We want to neutralize 2 ml of 1M HCl or NaOH. So the number of moles of HCl or NaOH that we want to neutralize are 0.002L X 1 mole/L = 0.002 moles. So, 50 ml of the buffer must contain 0.002 moles of the acid or base to neutralize NaOH and HCL respectively. Therefore, 100 ml of the buffer must contain 0.004 moles. We already know the molar ratio of acid and its conjugate base needed to achieve the required pH.
For pH 9,
0.004 moles of ammonia / x moles of ammonium chloride = 0.575 / 1
The number of moles of ammonium chloride required is 0.007. So, the final ratio of ammonia : ammonium chloride becomes 0.004 : 0.007.
4 ml of 1 M ammonia solution should be mixed with 0.374 g of ammonium chloride (molar mass = 53.491 g/mol) and the aqueous solution can be made up to 100 ml.
For pH 7,
0.004 moles of sodium phosphate dibasic / x moles of sodium phosphate monobasic = 0.617 / 1
The number of moles of sodium phosphate monobasic required is 0.0065. So, the final ratio of sodium phosphate dibasic : sodium phosphate monobasic becomes 0.004 : 0.0065.
0.568 g of sodium phosphate dibasic (molar mass = 141.96 g/mol) should be added to 0.78 g of sodium phosphate monobasic (molar mass = 119.98 g/mol) and the aqueous solution can be made up to 100 ml.
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