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

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.

the answer is   c.) 1575 j.

explanation: