If you have 200 ml of a 0.1M solution of acetic acid (pka=4.75),
1. How many ml of a solution of 1.O M NaOH would you need to adjust the pH to 5.2?
2. How many ml of water would you then add to make the solution of 0.05M in total acetic acid (ie., acetic acid + acetate ion)
200 mL x 0.1M = 20 mmoles.
1.
............HAc + OH^- ==> Ac^- H2O
initial.....20.0...0........0
added..............x...........
change.......-x....-x........+x
equil......20-x.....0.........+x
Substitute into the Henderson-Hasselbalch equation and solve for x which will be mmoles NaOH. You can take it from there.
2.
You have 200 mL of 0.1M = 20 mmoles to start. If you want it to be 0.05, wouldn't you dilute to 400 mL?
To answer these questions, we need to understand the concept of buffer solutions and the Henderson-Hasselbalch equation.
1. First, let's determine the initial concentration of acetic acid (CH3COOH) in the 200 ml solution.
- We are given that the initial concentration is 0.1 M.
- So, the initial number of moles of acetic acid in the solution = concentration x volume = 0.1 x 0.2 = 0.02 moles.
Next, let's calculate the required concentration of acetic acid (CH3COOH) at pH 5.2.
- The Henderson-Hasselbalch equation is pH = pKa + log(base/acid).
- Rearranging the equation, we have log(base/acid) = pH - pKa.
- Substituting the given values, we get log(base/acid) = 5.2 - 4.75 = 0.45.
- From logarithmic properties, we know that if log(base/acid) = x, then base/acid = 10^x.
- Therefore, base/acid = 10^0.45 = 3.548.
Since acetic acid and sodium hydroxide react in a 1:1 molar ratio, we can determine the required concentration of sodium hydroxide (NaOH) needed to adjust the pH to 5.2.
- The required number of moles of NaOH = base = 3.548 moles.
Now, let's calculate the volume of the 1.0 M NaOH solution needed.
- Volume (in liters) = moles/concentration = 3.548/1.0 = 3.548 L.
- Since 1 L = 1000 ml, the required volume of 1.0 M NaOH solution = 3.548 x 1000 = 3548 ml.
Therefore, to adjust the pH of the acetic acid solution to 5.2, you would need to add 3548 ml of a 1.0 M NaOH solution.
2. To make the solution 0.05 M in total acetic acid (acetic acid + acetate ion), we need to calculate the volume of water to add.
First, let's calculate the number of moles of acetic acid and acetate ion in the original 200 ml solution.
- Number of moles of acetic acid = initial concentration x volume = 0.1 x 0.2 = 0.02 moles.
- Number of moles of acetate ion = 0.02 moles (since acetic acid and acetate ion are formed in a 1:1 ratio).
To dilute the solution to a final concentration of 0.05 M, we need to calculate the total volume of the solution.
- The final concentration can be calculated using the formula: concentration = moles/volume.
- Rearranging the equation, we have volume = moles/concentration.
- Volume (in liters) = 0.02/0.05 = 0.4 L.
- Since 1 L = 1000 ml, the total volume of the solution = 0.4 x 1000 = 400 ml.
Therefore, to make the solution 0.05 M in total acetic acid, you would need to add (400 - 200) = 200 ml of water.