Question

50.0 mL of 0.10 M HA, a weak acid (Ka= 1.5 × 10-6), is titrated with 0.10 M B, a strong base. What is the pH at Vb= ½ Ve= 25.0 mL?

Answer 1

a)

pH at half Equivalence point

Note that this is a very special point, since this will occur:

strong base + weak acid = water + weak acid's conjugate

now.... recall that

for HA <-> H+ + A-

in the HALF equivalenc epoint

we have reacted HALF of th acid

so, the other HALF formed the conjugate base of the weak acid

i.e.

HA : A- ratio is 1:1

therefore,

according to henderson hasselbach equations

pH = pKa + log(A-/HA)

for A- = HA --> 1=1

then

log(1) = 0

therefore

pH = pKa, which we stated was special point, and it is ... since pKa = pH

now, calculate pKa

pKA = -log(Ka) = -log(1.5*10^-6) = 5.8239

pKa = 5.8239

pH = 5.8239