What is the pH of a sodium acetate solution prepared by adding 0.820g of sodium acetate to 100.0mL of water at 25 degrees?
To determine the pH of a sodium acetate solution, we need to consider its dissociation in water and calculate the concentration of the resulting ions. Sodium acetate (NaC2H3O2) dissociates completely in water into sodium ions (Na+) and acetate ions (C2H3O2-).
Step 1: Calculate the number of moles of sodium acetate:
Molar mass of sodium acetate (NaC2H3O2) = 82.03 g/mol
Number of moles = mass / molar mass
Number of moles = 0.820 g / 82.03 g/mol
Step 2: Calculate the molarity of the sodium acetate solution:
Molarity (M) = moles / volume (in liters)
Volume = 100.0 mL = 100.0/1000 = 0.100 L
Molarity (M) = (0.820 g / 82.03 g/mol) / 0.100 L
Step 3: Determine the concentration of acetate ions:
Since sodium acetate dissociates completely, the concentration of acetate ions will be equal to the molarity of the sodium acetate solution.
Step 4: Calculate the pOH of the solution:
pOH = -log10(Concentration of acetate ions)
Now, let's calculate the pOH:
pOH = -log10(0.820 g / 82.03 g/mol / 0.100 L)
Step 5: Calculate the pH of the solution:
pH + pOH = 14 (at 25 degrees)
Let's find the pH:
pH = 14 - pOH
Now let's substitute the pOH value to find the pH of the sodium acetate solution.
To determine the pH of a sodium acetate solution, we need to consider the dissociation of sodium acetate in water. Sodium acetate is formed by the combination of a sodium ion (Na+) and an acetate ion (CH3COO-). When sodium acetate dissolves in water, it dissociates into its constituent ions.
The dissociation reaction is as follows:
NaCH3COO(s) → Na+(aq) + CH3COO-(aq)
Since acetate ions are the conjugate base of acetic acid, they can react with water to form hydronium ions (H3O+) and acetate ions (CH3COO-). This is a weak acid-base reaction and leads to a slightly basic pH.
To calculate the pH, we need to consider the equilibrium constant for this reaction, which is known as the dissociation constant (Ka). The Ka expression for acetic acid is given by:
Ka = [H3O+][CH3COO-] / [CH3COOH]
In this case, because we are dealing with a salt (sodium acetate) dissolved in water, the initial concentration of acetic acid can be considered negligible. Thus, we can simplify the expression to:
Ka = [H3O+][CH3COO-]
Since the concentration of Na+ does not contribute to the acidity/basicity, we can ignore it in this calculation.
Next, we need to determine the concentration of the acetate ion (CH3COO-) in the solution. To do this, we need to convert the mass of sodium acetate to moles and then divide by the total volume of the solution in liters.
1. Convert mass to moles:
moles of sodium acetate = mass / molar mass
The molar mass of sodium acetate (NaCH3COO) is:
22.99 g/mol (Na) + 12.01 g/mol (C) + 3(1.01 g/mol) (H) + 12.01 g/mol (C) + 16.00 g/mol (O) + 16.00 g/mol (O) = 82.03 g/mol
moles of sodium acetate = 0.820 g / 82.03 g/mol = 0.00998 mol
2. Convert volume to liters:
volume = 100.0 mL = 0.100 L (since 1 L = 1000 mL)
3. Calculate the concentration of the acetate ion (CH3COO-):
concentration = moles / volume
concentration = 0.00998 mol / 0.100 L = 0.0998 M
Now, we can substitute this concentration into the Ka expression and solve for the H3O+ concentration. However, since the Ka value for acetic acid is relatively small (1.8 x 10^-5), we can assume that the concentration of H3O+ is equal to the concentration of CH3COO-.
[H3O+] = [CH3COO-] = 0.0998 M
To calculate the pH, we can use the formula:
pH = -log[H3O+]
Substituting the value of [H3O+] into the equation:
pH = -log(0.0998) ≈ 1.00
Therefore, the pH of the sodium acetate solution prepared by adding 0.820g of sodium acetate to 100.0mL of water at 25 degrees is approximately 1.00.