A buffer is prepared with benzoic acid, C6H5COOH 0.450 M, and its salt, C6H5COOK 0.350 M. The equilibrium equation is: C
Posted: Sun May 22, 2022 3:54 pm
A buffer is prepared with benzoic acid, C6H5COOH 0.450 M, and
its salt, C6H5COOK 0.350 M. The equilibrium equation is:
C6H5COOH(aq) + H2O(l) ↔ C6H5COO- + H3O+ Ka = 6.30 x
10-5
The mathematical expression to calculate the pH of said buffer
using the Henderson-Hasselbach equation is as follows:
Select one:
pH = 4.20 + log(0.450/0.350)
pH = 9.80 + log(0.450/0.350)
pH = 4.20 + log(0.350/0.450)
pH = 9.80 + log(0.350/0.450)
its salt, C6H5COOK 0.350 M. The equilibrium equation is:
C6H5COOH(aq) + H2O(l) ↔ C6H5COO- + H3O+ Ka = 6.30 x
10-5
The mathematical expression to calculate the pH of said buffer
using the Henderson-Hasselbach equation is as follows:
Select one:
pH = 4.20 + log(0.450/0.350)
pH = 9.80 + log(0.450/0.350)
pH = 4.20 + log(0.350/0.450)
pH = 9.80 + log(0.350/0.450)