You have a solution of acetic acid that has a K a of 3.5 × 10 –8 . If the concentration of the acetic acid were 2.40 M, what would be the concentration of H + at equilibrium?

2.90×10¯⁴M

Explanation:

Step 1:

Data obtained from the question.

Equilibrium constant for the acid , Ka = 3.5×10^–8

concentration of the acetic acid, [CH3COOH] = 2.40M

Concentration of Hydrogen ion, [H+] =..?

Step 2:

The balanced equation for the reaction.

CH3COOH(aq) <=> H+(aq) + CH3COO-(aq)

Step 3:

Determination of the concentration of Hydrogen ion, [H+]. This can be obtained as follow:

Initial concentration:

[CH3COOH] = 2.40M

[H+] = 0

[CH3COO-] = 0

During reaction:

[CH3COOH] = –y

[H+] = +y

[CH3COO-] = +y

At Equilibrium:

[CH3COOH] = 2.40 –y

[H+] = y

[CH3COO-] = y

Now, we can obtain the concentration of Hydrogen ion, H+ as follow:

Ka = [H+] • [CH3COO-] /[CH3COOH]

3.5×10^–8 = y×y/2.40

Cross multiply

y² = 3.5×10^–8 × 2.40

Take the square root of both side

y = √(3.5×10^–8 × 2.40)

y = 2.90×10¯⁴

[H+] = y = 2.90×10¯⁴M

Therefore, the concentration of H+ at equilibrium is 2.90×10¯⁴M

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.