matic reaction takes place in a 10-mL (0.010 L) solution that has a 13. An enzy
ID: 103490 • Letter: M
Question
matic reaction takes place in a 10-mL (0.010 L) solution that has a 13. An enzy phosphate concentration of 120 mM (0.120 M) and an inital pH of 7.00. During the reaction, 0.2 milliequivalents of acid (0.00020 mole of protons) are produced. Calculate the final pH of the solution. What would the final pH of the solution be in the absence of the phosphate buffer assuming that the other components of the solution have no significant buffering capacity and that the solution is initially at pH 7.00? (pKa's are given in problem 11B above; you do not need but one of these three values; you must choose the correct one.)Explanation / Answer
Since initial pH of solution is given as 7.00 we need to use pKa close to initial pH
Therefore pKa 6.86 need to be used in calculating final pH of the buffered solution
Accordingly we need to use the following equilibrium for the acid produced
H2PO4- <=====> HPO4-2 + H+
Initial concentration of H2PO4- = 0.120 M
Number of moles of acid, H+, produced = 0.00020
Volume of solution = 0.010 L
Concentration of produced acid, [H+] = 0.0002/0.010 = 0.020 M
From the above equation concentration of HPO4-2 is equal to that of acid produced = 0.020 M
Therefore final concentration of H2PO4- = 0.120 M – 0.020 M = 0.100 M
According to Henderson Hasselbach equation,
pH = pKa + log [A-]/[ HA] for a given acid dissociation as shown below
HA <====> A- + H+
Accordingly for the given phosphate solution with pKa 6.86 we need to substitute the above values into the equation to obtain the final pH of solution after the dissociation as shown below.
pH = pKa + log [HPO4-2]/[ H2PO4-]
pH = 6.86 + log [0.02]/[0.1]
= 6.86 + log [0.2]
= 6.86 + (-0.69897)
= 6.16103
= 6.16
Thus the final pH of solution after the given dissociation = 6.16
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