Calculate the ph at the equivalence point in titrating 0.100 m solutions of each of the following with 0.080 m naoh:

(a) hbr (aq), (b) hclo2 (aq), (c) benzoic acid (c6h5co2h).

(a) 7.0 (b) 7.30 (c) 8.42

Explanation:

(a) HBr is a strong acid and NaOH is also a very strong base, hence, the solution will be neutral give a pH of 7.0

(b) HClO3

ClO2- + H20 <=> HClO2- + OH-

ClO2- = 0.1*0.08/0.18 = 0.044M

Kb=Kw/Ka=1.0 x 10^-14/1.1 x 10^-2=9.09x10^-13

[OH-] = 2.0 x 10^-7M

p[OH] = 6.7

pH = 7.3

(c) C6H5COOH

C6H5COO- + H20 <=> C6H5COOH + OH-

[C6H5COO-] = 0.1 * 0.80/0.180 = 0.44

Kb=Kw/Ka=1.0 x 10^-14/6.3 x 10-²=1.59x10^-10

[OH] = 1.59x10-10/0.44 = 2.69x10-6

pOH = 5.58

pH = 8.42