make 200.0 mL of a pH 5.00 buffer from 0.200 M acetic acid and 0.100 M sodium acetate. Calculate the volume of 0.200 M acetic acid and the volume of 0.100 M sodium acetate required to make 200.0 mL of a buffer solution with pH = 5.00.
The Ka for acetic acid at 298 K is 2.00 x 10-5.
Let x = mL base
pH = pKa + log(base)/(acid)
5.00 = 4.699 + log(0.1x)/(200-x)*0.2
Solve for x and 200-x
Why did the acetic acid and sodium acetate go to the pH 5.00 buffer party? Because they wanted to be a good buffer and maintain a stable pH! Now, let's calculate their volumes to make sure they have a great time.
To begin, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the ratio of the concentrations of the acid and its conjugate base. The equation is as follows:
pH = pKa + log([A-]/[HA])
Since we want a pH of 5.00 and the pKa is given as 5.70, we can rearrange the equation to find the ratio [A-]/[HA]:
5.00 = 5.70 + log([A-]/[HA])
Now, let's convert the given molarity into moles per liter:
0.200 M acetic acid = 0.200 mol/L acetic acid
0.100 M sodium acetate = 0.100 mol/L sodium acetate
Since the total volume of the buffer is 200.0 mL, let's convert it to liters:
200.0 mL = 0.200 L
Now, let's use the ratio equation to find the moles of acetic acid and sodium acetate required to make the buffer solution:
[A-]/[HA] = 10^(pH - pKa)
[A-]/[HA] = 10^(5.00 - 5.70)
[A-]/[HA] = 10^(-0.70)
[A-]/[HA] = 0.1995
Now, let's set up the equations to calculate the volumes:
0.1995 = ([A-] moles / volume of A-) / ([HA] moles / volume of HA)
Knowing that the volume of the buffer solution is 0.200 L, we can rewrite the equations:
0.1995 = ([A-] moles / volume of A-) / ([HA] moles / (0.200 - volume of A-))
Solving for the unknowns, we can find the volumes of acetic acid and sodium acetate required to make the buffer solution.
Now, let's put our calculations into action and make that pH 5.00 buffer rock the party!
To calculate the volume of 0.200 M acetic acid and the volume of 0.100 M sodium acetate required to make a 200.0 mL buffer solution with pH = 5.00, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Given:
pH = 5.00
pKa = -log(Ka) = -log(2.00 x 10^-5) = 4.70
We can rearrange the Henderson-Hasselbalch equation to solve for the ratio of [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa)
[A-]/[HA] = 10^(5.00 - 4.70) = 1.995
Since the acetic acid (HA) and sodium acetate (A-) are in an equimolar ratio, we can assume that the volume of the acetic acid solution will be the same as the sodium acetate solution.
Let the volume of each solution required be x mL.
From the concentration, we can calculate the number of moles using the formula:
moles = concentration × volume
For acetic acid:
0.200 M × x mL = 1.995 × 0.100 M × x mL
0.200 = 1.995 × 0.100
0.200 = 0.1995
This equation is not satisfied, which means that a pH of 5.00 is not attainable with the given concentrations of acetic acid and sodium acetate.
To create a pH 5.00 buffer solution from acetic acid and sodium acetate, you will need to adjust the concentrations or find new concentrations that satisfy the equation.
To calculate the volume of 0.200 M acetic acid and the volume of 0.100 M sodium acetate required to make a pH 5.00 buffer, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
In this case, acetic acid (HA) is the weak acid, and sodium acetate (A-) is its conjugate base.
Given:
pH = 5.00
Ka = 2.00 x 10^-5
Step 1: Calculate the pKa
pKa = -log10(Ka)
pKa = -log10(2.00 x 10^-5)
pKa = 4.70
Step 2: Determine the ratio of [A-]/[HA]
pH = pKa + log ([A-]/[HA])
5.00 = 4.70 + log ([A-]/[HA])
To create a buffer solution with pH 5.00, we want to have an equal concentration of acetic acid and acetate ions. Therefore, the ratio [A-]/[HA] should be 1:1.
Step 3: Calculate the volume of acetic acid
To find the volume of acetic acid needed, we can use the formula:
moles of acetic acid (HA) = concentration of acetic acid (mol/L) x volume of acetic acid (L)
We are given that the concentration of acetic acid is 0.200 M and we need to make 200.0 mL of the buffer solution.
moles of acetic acid (HA) = 0.200 mol/L x 0.200 L
moles of acetic acid (HA) = 0.040 mol
Step 4: Calculate the volume of sodium acetate
Since the ratio of [A-] to [HA] is 1:1, the moles of sodium acetate (A-) required will be the same as the moles of acetic acid (HA):
moles of sodium acetate (A-) = 0.040 mol
Next, we can calculate the volume of 0.100 M sodium acetate required using the formula:
volume of sodium acetate (L) = moles of sodium acetate (mol) / concentration of sodium acetate (mol/L)
volume of sodium acetate (L) = 0.040 mol / 0.100 mol/L
volume of sodium acetate (L) = 0.4 L = 400 mL
Therefore, you would need 400 mL of 0.100 M sodium acetate and 200 mL of 0.200 M acetic acid to make a 200.0 mL buffer solution with a pH of 5.00.