1). Calculate the initial and final pH when 5.0, 10.0, and 100mL of 0.100M HCL is added to
(a) 100mL of water
(b) 100mL of a buffer which is 1.50M acetic acid and 1.20M sodium acetate
2). Develop a recipe for making this buffer from acetic acid, sodium hydroxide, and water
1a. 5 mL added.
(HCl) = 0.1M x 5 mL/105 mL = ?. Convert to pH.
10 mL and 100 mL are done the same way.
1b. millimols HCl added = 5 x 0.1 =0.5
mmols HAc = mL x M = about 150
mmols Ac&- = about 120
.......Ac^- + H^+ ==> HAc
I......120....0........150
add..........0.5..........
C.....-0.5..-0.5.......+0.5
E......119.5..0........150.5
Substitute the E line into the Henderson-Hasselbalch equation and solve for pH.
The 10 and 100 mL are done the same way.
Dr.Bob222, thank you for the assistance with 1a and 1b, however, I'm kind of confused with 2! What does recipe entail? A description? A series of calculations?
To calculate the pH in the provided scenarios, we need to consider the concentrations, volumes, and pKa values of the acid-base systems involved. In this case, we are working with hydrochloric acid (HCl) as the strong acid and acetic acid (CH3COOH) as the weak acid.
1) Calculating the pH when HCl is added to:
(a) 100mL of water:
Since HCl is a strong acid, it dissociates completely in water. Therefore, we can calculate the initial and final pH using the concentration of HCl.
Initial pH:
The initial pH can be calculated using the equation: pH = -log[H+].
[HCl] = 0.100M (given)
[H+] = 0.100M
pH = -log(0.100) = 1.00
Final pH:
When HCl is added to water, the final volume becomes 100mL + 5.0mL or 10.0mL or 100mL. However, since HCl is a strong acid, it will completely dissociate, providing additional H+ ions in the solution. Therefore, the concentration of H+ is simply the sum of the initial concentration of HCl and the additional H+ concentration provided by HCl.
[H+] = [HCl] (initial concentration of HCl) + [HCl] (additional H+ concentration provided by HCl)
[H+] = 0.100M + (0.100M / (100mL + added volume))
Substituting the added volumes:
For 5.0mL HCl:
[H+] = 0.100M + (0.100M / (100mL + 5.0mL)) = 0.1044M
pH = -log(0.1044) = 0.98
For 10.0mL HCl:
[H+] = 0.100M + (0.100M / (100mL + 10.0mL)) = 0.1053M
pH = -log(0.1053) = 0.98
For 100mL HCl:
[H+] = 0.100M + (0.100M / (100mL + 100mL)) = 0.150M
pH = -log(0.150) = 0.82
(b) 100mL of a buffer which is 1.50M acetic acid and 1.20M sodium acetate:
In this scenario, we are dealing with a buffer system comprising acetic acid (weak acid) and its conjugate base sodium acetate (salt). To calculate the pH, we need to take into account the Henderson-Hasselbalch equation:
pH = pKa + log([A-] / [HA])
Where pKa is the acid dissociation constant of acetic acid, [A-] is the concentration of the acetate ion, and [HA] is the concentration of acetic acid.
The pKa of acetic acid is 4.76.
Initial pH:
Since there is no addition of HCl, the initial pH of the buffer solution can be calculated using the Henderson-Hasselbalch equation with the provided concentrations.
pH = 4.76 + log(1.20M / 1.50M) = 4.76 - log(0.80) = approximately 4.58
Final pH:
When HCl is added to the buffer solution, it reacts with the acetate ion to form more acetic acid, disturbing the buffer system and changing the pH. To calculate the final pH, we need to determine the new concentrations of the weak acid and its conjugate base.
For 5.0mL HCl:
[H+] = 0.100M + (0.100M / (100mL + 5.0mL)) = 0.1044M
New concentration of acetate ion (A-):
[A-] = initial concentration - [HCl] (additional H+ concentration provided by HCl)
[A-] = 1.20M - 0.1044M = 1.0956M
New concentration of acetic acid (HA):
[HA] = initial concentration + [HCl] (additional H+ concentration provided by HCl)
[HA] = 1.50M + 0.1044M = 1.6044M
Calculating the final pH using the Henderson-Hasselbalch equation:
pH = 4.76 + log(1.0956M / 1.6044M) = 4.76 - log(0.6830) = approximately 4.98
Similarly, you can calculate the final pH for 10.0mL and 100mL of HCl by substituting the corresponding values for [H+] in the Henderson-Hasselbalch equation.
2) To develop a recipe for making the buffer from acetic acid, sodium hydroxide, and water, we need to determine the required concentrations and volumes.
Given that we want to prepare a buffer with final concentrations of 1.50M acetic acid and 1.20M sodium acetate, we can use the Henderson-Hasselbalch equation to find the required ratio of [A-] to [HA].
Using the equation:
pH = pKa + log([A-] / [HA])
We can substitute the desired pH and pKa value of acetic acid to solve for the concentration ratio:
4.76 = log([A-] / [HA])
Taking the antilog of both sides, we have:
10^4.76 = [A-] / [HA]
Simplifying, we find that:
[A-] / [HA] = approximately 2.51
Next, we need to determine the total volume of the buffer solution. This can be any desired volume, but for this example, let's assume a total volume of 1 liter (1000mL).
To calculate the required concentration and volumes for preparing the buffer:
1) Determine the volume of the acetic acid solution:
Assume we will use x mL of acetic acid solution.
The concentration of acetic acid in the solution is given as 1.50M.
Therefore, the moles of acetic acid can be calculated as:
Moles of acetic acid = 1.50M x (x/1000)
2) Determine the volume of the sodium acetate solution:
Since we want the ratio of [A-] to [HA] to be 2.51, we can set up the equation:
1.20M x (1000 - x) = 2.51 x (1.50M x (x/1000))
This equation ensures that the concentration ratio in the final solution matches the desired buffer pH.
Solving this equation will give you the volume of the sodium acetate solution, which is (1000 - x) mL.
3) The remaining volume of the buffer solution will be water, which is simply 1000mL - (volume of acetic acid solution + volume of sodium acetate solution).
By following these calculations, you can determine the specific amounts of acetic acid, sodium acetate, and water needed to prepare the buffer solution.