Question

1) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH).

12345 100.0 mL of 0.100 M HC₃H₅O₂ (K_a = 1.3 x 10^-5) by 0.100 M NaOH

12345 100.0 mL of 0.100 M HCl by 0.100 M NaOH

12345 100.0 mL of 0.100 M KOH by 0.100 M HCl

12345 100.0 mL of 0.100 M NH₃ (K_b = 1.8 x1 0^-5) by 0.100 M HCl

12345 100.0 mL of 0.100 M HNO₂ (K_a = 4.0 x 10^-4) by 0.100 M NaOH

Answer 1

100.0 mL of 0.100 M HC3H5O2 (Ka = 1.3 x 10-5) by 0.100 M NaOH

this forms a buffer, therefore,

pH = pKa + log(conjugate/acid)

pH = 4.75 + log(0.1/0.1) ) 4.75

2)

halfway, this will have a very low pH i.e. 3

3)

this is basic, since KOH is strong base

4)

NH3 forms a buffer so

pOH = Kb + log(conjugate/base) = 4,75

pH = 9.25

5)

this will form a buffer similar to (1)

then

pH = pKa + log(conjugate/acid)

since in halfway conjugate= acid then

pH = pKa = -log(4*10^-4) = 3.3979

THEN

most acidic:

Solution (2)

Solution/buffer (5)

Soluition/bufer (1)

basic...

Solution/buffer (4)

Basic Solution in (3)

Most basic