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

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.