Calculate the pH of a 0.02 M solution of ascorbic acid ( K a1 = 7.9 × 10 –5; K a2 is 1.6 × 10 –12).
1 answer:
You might be interested in
Answer:
pH = 2.46
Explanation:
Hello there!
In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:
Whereas the moles of the salt are computed as shown below:
So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:
Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:
Whose equilibrium expression is:
Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:
Whereas x is:
Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:
Regards!