A 25.0 mL sample of 0.150 M hypochlorous acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The Kaof hypochlorous acid is 3.0x10^-8. a) 10.20
b) 7.00
c) 6.48
d) 7.52
e) 14.52

pH = 10.20

Explanation:

The HClO reacts with NaOH as follows:

HClO + NaOH → H2O + NaClO

Where HClO and NaOH react in a 1:1 reaction.

As the concentration of both reactions is the same and the reaction is 1:1, to reach equivalence point are required the same 25.0mL.

And the NaClO produced decreases its concentration in 2 because the volume is doubled.

The concentration of NaClO is: 0.150M / 2 = 0.075M

The equilibrium of NaClO is:

NaClO(aq) + H2O(l) ⇄ HClO(aq) + OH-(aq)

Where Kb of reaction is 1.0x10⁻¹⁴ / Ka =

1.0x10⁻¹⁴ / 3.0x10⁻⁸ = 3.33x10⁻⁷ = [HClO] [OH-] / [NaClO]

[NaClO] = 0.075M

As both HClO and OH- comes from the same equilibrium,

[HClO] = [OH-] = X

Where X is the reactoin coordinate

Replacing:

3.33x10⁻⁷ = [X] [X] / [0.075M]

2.5x10⁻⁸ = X²

X = 1.58x10⁻⁴M = [OH-]

pOH = -log [OH-]

pOH = 3.80

pH = 14 - pOH

answer: your mm

explanation:

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.