Question

определить молярную концентрацию CH3COOH в водном растворе ph которого равен 3, степень диссоциации кислоты равна 1%

Answer 1

1)pH= - lg[H+]
3= - lg[H+]
[H+]= 10⁻³ 
2) [H+]= α*C
C= [H+] / α=10⁻³ / 0.01= 0.1 моль/л

Answer 2

Calculation of Molar Concentration of CH3COOH in a Solution with pH 3 and 1% Degree of Dissociation

To determine the molar concentration of CH3COOH in a water solution with a pH of 3 and a degree of dissociation of 1%, we need to consider the dissociation reaction of acetic acid (CH3COOH) in water:

CH3COOH ⇌ CH3COO- + H+

The degree of dissociation (α) is defined as the ratio of the concentration of dissociated molecules to the initial concentration of the acid. In this case, the degree of dissociation is given as 1%, which means that only 1% of the acetic acid molecules dissociate into CH3COO- and H+ ions.

Let's assume the initial concentration of CH3COOH is C (in mol/L). Since the degree of dissociation is 1%, the concentration of CH3COO- and H+ ions can be calculated as 0.01C (0.01 times the initial concentration).

According to the law of mass action, the equilibrium constant (Ka) for the dissociation reaction can be expressed as:

Ka = [CH3COO-][H+] / [CH3COOH]

At equilibrium, the concentration of CH3COO- and H+ ions will be 0.01C, and the concentration of undissociated CH3COOH will be (1 - 0.01)C = 0.99C.

Given that the pH of the solution is 3, we can use the relationship between pH and the concentration of H+ ions:

pH = -log[H+]

From this equation, we can calculate the concentration of H+ ions as 10^(-pH).

Now, we can substitute the concentrations into the equilibrium constant expression:

Ka = (0.01C)(10^(-pH)) / (0.99C)

Simplifying the equation:

Ka = 0.01 / (0.99 * 10^pH)

The equilibrium constant (Ka) for acetic acid is approximately 1.8 * 10^(-5) at 25°C.

Setting the equilibrium constant expression equal to the known Ka value:

1.8 * 10^(-5) = 0.01 / (0.99 * 10^pH)

Solving for C, the initial concentration of CH3COOH:

C = (0.01 * 10^pH) / (0.99 * 1.8 * 10^(-5))

Now, we can substitute the given pH value of 3 into the equation to calculate the molar concentration of CH3COOH:

C = (0.01 * 10^3) / (0.99 * 1.8 * 10^(-5))

C ≈ 5.56 mol/L

Therefore, the molar concentration of CH3COOH in the water solution with a pH of 3 and a 1% degree of dissociation is approximately 5.56 mol/L.

Please note that this calculation assumes ideal behavior and neglects any other factors that may affect the dissociation of acetic acid.