Calculate the ph of a 1L solution to wgich has been added 25ml of 10mM acetic acid and 25ml of 30mM sodium acetate
To calculate the pH of the solution, we need to determine the concentrations of acetic acid and acetate in the final solution.
Step 1: Convert the volume of acetic acid and sodium acetate to liters:
- Volume of acetic acid = 25 mL = 25/1000 L = 0.025 L
- Volume of sodium acetate = 25 mL = 25/1000 L = 0.025 L
Step 2: Calculate the number of moles of acetic acid and sodium acetate:
- Moles of acetic acid = concentration × volume = 10 mM × 0.025 L = 0.25 mmol = 0.00025 mol
- Moles of sodium acetate = concentration × volume = 30 mM × 0.025 L = 0.75 mmol = 0.00075 mol
Step 3: Calculate the total moles of acetic acid and sodium acetate:
- Total moles = moles of acetic acid + moles of sodium acetate
- Total moles = 0.00025 mol + 0.00075 mol = 0.001 mol
Step 4: Calculate the molar concentrations of acetic acid and acetate:
- Molar concentration of acetic acid = moles/volume = 0.001 mol / 1 L = 0.001 M
- Molar concentration of acetate = moles/volume = 0.001 mol / 1 L = 0.001 M
Step 5: Use the Henderson-Hasselbalch equation to calculate the pH:
- pH = pKa + log([Acetate]/[Acetic acid])
- The pKa value of acetic acid is 4.76.
Substituting the values:
- pH = 4.76 + log(0.001 M / 0.001 M)
- pH = 4.76 + log(1)
- pH = 4.76 + 0
- pH = 4.76
Therefore, the pH of the solution is 4.76.
To calculate the pH of the solution, we need to take into account the concentration of acetic acid, sodium acetate, and the equilibrium between them. Acetic acid (CH3COOH) and sodium acetate (CH3COONa) form a buffer system, which means they can resist changes in pH when small amounts of acid or base are added.
The Henderson-Hasselbalch equation can be used to calculate the pH of a buffer solution:
pH = pKa + log ([A-] / [HA])
Where:
- pH is the desired pH of the solution.
- pKa is the logarithmic value of the acid dissociation constant of acetic acid. The pKa of acetic acid is 4.74.
- [A-] is the concentration of the conjugate base (acetate) in the solution.
- [HA] is the concentration of the weak acid (acetic acid) in the solution.
We need to calculate the concentrations [A-] and [HA] using the given information.
1. Concentration of acetic acid (CH3COOH):
- Volume of acetic acid = 25 mL = 0.025 L
- Concentration of acetic acid = 10 mM = 0.01 mol/L
- [HA] = concentration of acetic acid = 0.01 mol/L
2. Concentration of sodium acetate (CH3COONa):
- Volume of sodium acetate = 25 mL = 0.025 L
- Concentration of sodium acetate = 30 mM = 0.03 mol/L
- [A-] = concentration of sodium acetate = 0.03 mol/L
Now, substitute the values into the Henderson-Hasselbalch equation:
pH = 4.74 + log (0.03 / 0.01)
Simplifying the equation:
pH = 4.74 + log (3)
Using a calculator, find the logarithm:
pH = 4.74 + 0.48
Calculating the sum:
pH ≈ 5.22
Therefore, the pH of the 1L solution containing 25 mL of 10 mM acetic acid and 25 mL of 30 mM sodium acetate is approximately 5.22.
acetic acid = HAc
sodium acetate = NaAc
The way I interpret the problem as to how the author wants it solved is that you have 1L of solution containing 25 mM HAc and 30 mM NaAc; however, technically that isn't what it says to me.
(HAc) = 0.01 M x (25/1000) = ?
(NaAc) =0.03 M x (25/1000) = ?
Then pH = pKa for HAc + [log(NaAc)/(HAc)]
Post your work if you get stuck.