To calculate the pH of a solution of a salt resulting from a weak acid and a strong base, we need to understand that the anion of the weak acid (in this case, acetate, CH3CO2-) will act as a base in water, accepting a proton (H+) to form the weak acid (acetic acid, HCH3CO2). This is a basic solution, and we can use the formula for the pOH of a basic solution and then convert to pH.

Here are the steps:

1. Write the equilibrium expression for the reaction of the acetate ion with water:

CH3CO2- + H2O ⇌ HCH3CO2 + OH-

2. Use the pKa of the acetic acid to find the Ka and then the Kb for the acetate ion. The relationship between Ka, Kb, and Kw (the ion product of water, 1.0 x 10^-14 at 25°C) is Ka x Kb = Kw.

Ka = 10^-pKa = 10^-4.76 = 1.74 x 10^-5

Kb = Kw / Ka = (1.0 x 10^-14) / (1.74 x 10^-5) = 5.75 x 10^-10

3. Use the Kb in an ICE table (Initial, Change, Equilibrium) to find the concentration of OH- ions. Since we start with a 0.67M solution of KCH3CO2, we start with 0.67M CH3CO2- ions. The change in concentration is -x for CH3CO2-, +x for OH-, and +x for HCH3CO2. At equilibrium, we have 0.67-x M CH3CO2-, x M OH-, and x M HCH3CO2.

Kb = [OH-][HCH3CO2] / [CH3CO2-] = x*x / (0.67-x) = 5.75 x 10^-10

Since Kb is very small, we can assume that x is very small compared to 0.67, so the equation simplifies to x^2 / 0.67 = 5.75 x 10^-10. Solving for x gives x = sqrt(0.67 * 5.75 x 10^-10) = 6.53 x 10^-6 M, which is the concentration of OH- ions.

4. Use the concentration of OH- ions to find the pOH, and then the pH. The pOH is -log[OH-], and the pH is 14 - pOH.

pOH = -log(6.53 x 10^-6) = 5.2

pH = 14 - 5.2 = 8.8

So, the pH of a 0.67M solution of potassium acetate at 25°C is approximately 8.8.

Question

To calculate the pH of a solution of a salt resulting from a weak acid and a strong base, we need to understand that the anion of the weak acid (in this case, acetate, CH3CO2-) will act as a base in water, accepting a proton (H+) to form the weak acid (acetic acid, HCH3CO2). This is a basic solution, and we can use the formula for the pOH of a basic solution and then convert to pH.

Here are the steps:

1. Write the equilibrium expression for the reaction of the acetate ion with water:

CH3CO2- + H2O ⇌ HCH3CO2 + OH-

2. Use the pKa of the acetic acid to find the Ka and then the Kb for the acetate ion. The relationship between Ka, Kb, and Kw (the ion product of water, 1.0 x 10^-14 at 25°C) is Ka x Kb = Kw.

Ka = 10^-pKa = 10^-4.76 = 1.74 x 10^-5

Kb = Kw / Ka = (1.0 x 10^-14) / (1.74 x 10^-5) = 5.75 x 10^-10

3. Use the Kb in an ICE table (Initial, Change, Equilibrium) to find the concentration of OH- ions. Since we start with a 0.67M solution of KCH3CO2, we start with 0.67M CH3CO2- ions. The change in concentration is -x for CH3CO2-, +x for OH-, and +x for HCH3CO2. At equilibrium, we have 0.67-x M CH3CO2-, x M OH-, and x M HCH3CO2.

Kb = [OH-][HCH3CO2] / [CH3CO2-] = x*x / (0.67-x) = 5.75 x 10^-10

Since Kb is very small, we can assume that x is very small compared to 0.67, so the equation simplifies to x^2 / 0.67 = 5.75 x 10^-10. Solving for x gives x = sqrt(0.67 * 5.75 x 10^-10) = 6.53 x 10^-6 M, which is the concentration of OH- ions.

4. Use the concentration of OH- ions to find the pOH, and then the pH. The pOH is -log[OH-], and the pH is 14 - pOH.

pOH = -log(6.53 x 10^-6) = 5.2

pH = 14 - 5.2 = 8.8

So, the pH of a 0.67M solution of potassium acetate at 25°C is approximately 8.8.

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