Surely you mean Ve = volume at the equivalence point.
The Henderson-Hasselbalch equation is
pH = pKa + log(base)/(acid)
If pH = pKa, then log B/A = 0 and B/A = 1
If pH = pKa + 1, then log B/A = 1 and B/A = 10
If pH = pKa -1, then log B/A = -1 and B/A = 0.1
..........HA + OH^- ==> A^- + H2O
initial....1.....0.......0......0
add.............x..............
change.....-x....-x......x.....x
equil.....1-x.....0......x.....x
The three cases above are done this way.
When B/A = 1, then [(x)/(1-x)] = 1
Solve for x and obtain x = 1-x so you are half way to the equivalence point.
When B/A = 10, then [(x)/(1-x)]=10
Solve for x = 0.90909 and that is the fraction to Ve.
When B/A = 0.1, then......
Set up an ICE chart and substitute moles into it for base and acid and use the HH equation to solve for pH for the specific points.
Post your work if you get stuck.
Consider the titration of the weak acid HA with NaOH. At what fraction of Ve does pH = pKa - 1? At what fraction of Ve does pH = pKa + 1? Calculate the pH at these two points plus Vb = 0, 1/2Ve, Ve and 1.2Ve, if 100 mL of 0.100 M anilinium bromide (pKa = 4.601) is titrated with 0.100 M NaOH. {Ve is Volume at equilibrium,Vb is volume of base}
2 answers
Ph8.3
2 answers