You have made a buffer solution where you added 9 mL 1M Acetic Acid and 0.82 grams Sodium Acetate and diluted it with deionized water to 100 mL. If the Ka for Acetic Acid is 1.76*10-5 and the molecular weight of Sodium Acetate is 82.03 g/mole, what is the pH of the buffer you prepared?
base = NaAc = (grams/molar mass)/L
acid = HAc = 1M x (9/100) = ?
Substitute into the Henderson-Hasselbalch equation and solve for pH.
Post your work if you get stuck.
To determine the pH of the buffer solution you prepared, we can start by calculating the concentrations of the acetic acid and acetate ion in the solution. This can be done using the following steps:
Step 1: Calculate the number of moles of acetic acid (CH3COOH):
Molar mass of acetic acid (CH3COOH) = 60.05 g/mole
Number of moles of acetic acid = (9 mL / 1000 mL) * (1 mole / 1 L) * (1 M) = 0.081 moles
Step 2: Calculate the concentration of acetic acid (CH3COOH):
Volume of the solution = 100 mL = 0.1 L
Concentration of acetic acid = (0.081 moles) / (0.1 L) = 0.81 M
Step 3: Calculate the concentration of acetate ion (CH3COO-):
Number of moles of sodium acetate (NaCH3COO) = (0.82 grams) / (82.03 g/mole) = 0.01 moles
Concentration of acetate ion = (0.01 moles) / (0.1 L) = 0.1 M
Now that we have the concentrations of acetic acid and acetate ion, we can determine the pH of the buffer using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where:
pH = the pH of the buffer solution
pKa = the negative logarithm (base 10) of the acid dissociation constant Ka
[A-] = concentration of the conjugate base (acetate ion)
[HA] = concentration of the acid (acetic acid)
Given that the pKa for acetic acid (CH3COOH) is 1.76 x 10^-5, and the concentrations of acetate ion and acetic acid are 0.1 M and 0.81 M, respectively, we can calculate the pH as follows:
pH = -log(1.76 x 10^-5) + log(0.1/0.81)
Calculating this expression using a calculator, the pH of the buffer solution you prepared is approximately 4.74.