Question

I have this confusing problem with a phosphate hypothesis, can you help?

Answer 1

The phosphate buffer has 2 components and a pH of around 7. One component absorbs acids and the other absorbs bases. The buffer should be capable of absorbing up to half its own 'weight' of excess acid or base.
Let's test this hypothesis.
To do so, we need to quantitatively compare the amounts of buffer and added acid and base in the experiment. We use units of milli-mole (one thousandth of a mole), written as 'mmol'.
The phosphate buffer had 2.5 mL of 0.1M Na2HPO4, the acid absorber, where 0.1M means a concentration of 0.1 moles per liter of solution. Convert this to mmol/mL as follows:
(0.1 Mole/L) x (1000 mmol / 1 Mole) x (1L / 1000 mL) x (2.5 mL) = 0 .25 mmol
The same amount of NaH2PO4 base absorber is present as well.
The amount of added acid or base is calculated from the number of drops, knowing that one drop is approximately 0.05 mL, and the concentrations of the acid and base was 0.5M, as follows:
(0.5 Mole/L) x (1000 mmol / 1 Mole) x (1L / 1000 mL) x (0.05 mL/drop) = 0.025 mmol / drop
(a) Look again at the graph of pH vs. drops of acid/base and decide at what number of drops the buffer failed to maintain a nearly constant pH.
(b) Record this pH value for the addition of both acid and base.
(c) Multiply the number of drops by the conversion factor above, and divide by the total volume of buffer used (5 mL) to obtain the total buffering capacity, in mmol/mL. Your answer should be written as follows:
Phosphate buffering capacity
= X mmol acid per mL of buffer
= Y mmol base per mL of buffer
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