At equivalence point, X mL of 0.02 M HCl is treated with 5 mL of 0.02 M weak base.

The pK_b of the weak base is 5.69.

The pH of the solution at half of the equivalence point is Y.

Find (X + Y).

Options:

1. 15
2. 8.81
3. 13.31
4. 3.81

Solution:

Moles of weak base:

= 0.02 × 5 × 10^-3

= 1 × 10^-4 mol

At equivalence:

Moles of HCl = Moles of base

0.02 × X/1000 = 1 × 10^-4

X = 5 mL

At half equivalence point:

pOH = pK_b

= 5.69

Therefore:

pH = 14 − 5.69

= 8.31

Thus:

X + Y

= 5 + 8.31

= 13.31

Answer:

13.31 (Option 3)