25.0 g of potassium acetate (CH 3 COOK) are dissolved in 1.00 L of water. Calculate the pH of this solution. (PKa = 4.76 for CH3COOH)
To calculate the pH of the solution, we need to consider the dissociation of potassium acetate.
Potassium acetate, CH3COOK, is the salt of a weak acid, CH3COOH, which means it will partially dissociate in water. The acetate ion (CH3COO-) is the conjugate base of the weak acid. This means it will react with water to form hydroxide ions (OH-) and the weak acid.
The dissociation equation for potassium acetate is:
CH3COOK --> CH3COO- + K+
Since it is a salt of the weak acid, the concentration of hydroxide ions formed will depend on the concentration of the acetate ion. To find the concentration of the acetate ion, we will assume that potassium acetate is completely dissociated in water.
Given:
Mass of potassium acetate (CH3COOK) = 25.0 g
Volume of water = 1.00 L
Step 1: Calculate the number of moles of potassium acetate.
Molar mass of CH3COOK (potassium acetate) = 98.14 g/mol
Number of moles = mass / molar mass
Number of moles of potassium acetate = 25.0 g / 98.14 g/mol
Step 2: Calculate the concentration of the acetate ion.
Since potassium acetate is completely dissociated in water, the concentration of the acetate ion is equal to the number of moles of potassium acetate divided by the volume of water (in liters).
Concentration of acetate ion = (Number of moles of potassium acetate) / (Volume of water)
Step 3: Calculate the pOH of the solution.
Since the acetate ion reacts with water to form hydroxide ions, we need to find the pOH of the solution. The pOH is calculated using the concentration of hydroxide ions (OH-).
pOH = -log10(OH- concentration)
Step 4: Calculate the pH of the solution.
The pH scale is related to the pOH by the equation: pH + pOH = 14
Therefore, pH = 14 - pOH
Now we can go ahead and calculate the pH of the solution:
Step 1: Calculate the number of moles of potassium acetate:
Number of moles of potassium acetate = 25.0 g / 98.14 g/mol = 0.2546 mol
Step 2: Calculate the concentration of the acetate ion:
Concentration of acetate ion = (0.2546 mol) / (1.00 L) = 0.2546 M
Step 3: Calculate the pOH of the solution:
pOH = -log10(Concentration of hydroxide ions)
Since the acetate ion reacts with water to form hydroxide ions, the concentration of hydroxide ions is equal to the concentration of the acetate ion.
pOH = -log10(0.2546 M)
Step 4: Calculate the pH of the solution:
pH = 14 - pOH
Now, substitute the values in Step 4 to calculate the pH.
To calculate the pH of a solution containing potassium acetate, we need to consider the dissociation of the acetic acid (CH3COOH) and the hydrolysis of acetate ion (CH3COO-) in water.
The dissociation of acetic acid can be represented by the equation:
CH3COOH ⇌ H+ + CH3COO-
Acetic acid partially dissociates in water, releasing both hydrogen ions (H+) and acetate ions (CH3COO-). The acetate ions can then undergo hydrolysis, reacting with water and form hydroxide ions (OH-). Therefore, the equation for the hydrolysis of acetate ions can be written as:
CH3COO- + H2O ⇌ CH3COOH + OH-
Since the concentration of the potassium acetate solution is given as 25.0 g in 1.00 L of water, we need to convert the mass of potassium acetate to moles. The molar mass of potassium acetate (CH3COOK) is calculated as follows:
Molar mass of CH3COOK = (12.01 g/mol x 2) + (1.01 g/mol x 3) + 39.10 g/mol + 16.00 g/mol = 98.15 g/mol
Therefore, the number of moles of potassium acetate is:
moles of CH3COOK = mass / molar mass = 25.0 g / 98.15 g/mol = 0.2545 mol
Since 1 mol of CH3COOK dissociates to 1 mol of CH3COO-, the concentration of acetate ions is also 0.2545 mol/L.
Now, let's consider the hydrolysis of acetate ions. Since CH3COO- reacts with water (H2O), it forms acetic acid (CH3COOH) and hydroxide ions (OH-). The reaction consumes equal moles of CH3COO- and produces OH- ions in the same concentration.
Therefore, the concentration of hydroxide ions (OH-) is also 0.2545 mol/L.
Now, we can calculate the pOH (-log10[OH-]) using the concentration of hydroxide ions:
pOH = -log10[OH-] = -log10(0.2545) ≈ 0.595
Now, to find the pH, we can use the relation:
pH + pOH = 14
Therefore:
pH = 14 - 0.595 ≈ 13.405
So, the pH of the solution is approximately 13.405.