80 mL of benzoic acid (C6H5COOH) of density d = 0.96 g / mL are mixed with water and the volume is brought to 540 mL. Calculate the pH of this solution, knowing that Ka = 6.3x10-5 mol / L.

Explanation:

Concentration of benzoic acid, [benzoic acid] =1M (initially)

concentration of benzoate, [benzoate] =1M (initially)

Let's consider V ml of benzoic acid be taken.

So (300−V) ml of sodium benzoate is taken.

∴ [benzoicacid]

final

=

300ml

1M×Vml

[benzoate]

final

=

300ml

1M×(300−V)ml

∴ pH=pK

a

+log

[benzoicacid]

final

[benzoate]

final

⇒4.5=4.2+log{

V/300

(300−V)/300

}

⇒

V

300−V

=10

0.3

≈2

⇒V=100ml

∴ 100ml of benzoic acid had been taken.

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.