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 acetate
Then 200-x = mL acid
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5.00 = 4.70 + log(0.1x)/[0.2(200-x)]
in which mL x M = millimoles so
0.1x is mmoles acetate and
0.2x(200-x) mmoles acid.
Solve for x = volume of base. I get 160 mL and 40 mL for the acid but check my work.
Thank you
where does the 4.70 come from?
How did u solve for X
Sarah, 4.70 is what you get when you do the -log (Ka)
To calculate the volumes of acetic acid and sodium acetate required to make a buffer solution with a pH of 5.00, we need to use the Henderson-Hasselbalch equation, which relates the pH, pKa (the negative logarithm of the acid dissociation constant), and the ratio of the concentrations of the conjugate acid and base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([conjugate base]/[acid])
In this case, acetic acid (CH3COOH) is the acid, and sodium acetate (CH3COONa) is the conjugate base. The pKa of acetic acid is given as 2.00 x 10^-5.
First, let's find the ratio of the concentrations of the conjugate base to the acid. We'll use the pH and pKa values in the Henderson-Hasselbalch equation:
5.00 = -log(2.00 x 10^-5) + log([CH3COONa]/[CH3COOH])
Simplifying the equation:
log([CH3COONa]/[CH3COOH]) = 5.00 - (-log(2.00 x 10^-5))
log([CH3COONa]/[CH3COOH]) = 5.00 + 5.30
log([CH3COONa]/[CH3COOH]) = 10.30
Now, to get the ratio of the concentrations, we exponentiate both sides of the equation:
[CH3COONa]/[CH3COOH] = 10^(10.30)
[CH3COONa]/[CH3COOH] = 1.9953 x 10^10
Now, let's calculate the volumes of acetic acid and sodium acetate required.
Volume of 0.200 M acetic acid:
We know the total volume of the buffer solution is 200.0 mL. Let's assume the volume of acetic acid is V1 mL. Then, the volume of sodium acetate would be (200.0 - V1) mL.
Using the concentration and volume, we can write the equation:
0.200 M x V1 mL = moles of acetic acid
We need to find the number of moles of acetic acid required. The formula for moles is:
moles of acetic acid = concentration x volume (in liters)
moles of acetic acid = 0.200 M x (V1/1000)L
Next, we need to convert moles of acetic acid to moles of sodium acetate using the molar ratio from the balanced chemical equation:
1 mole of acetic acid reacts with 1 mole of sodium acetate
Since the ratio of the concentrations was calculated to be 1.9953 x 10^10, we can use this ratio to find the moles of sodium acetate:
moles of sodium acetate = ratio x moles of acetic acid
moles of sodium acetate = (1.9953 x 10^10) x (0.200 M x (V1/1000)L)
Now, let's calculate the volume of 0.100 M sodium acetate:
0.100 M x (200.0 - V1) mL = moles of sodium acetate
Similar to the previous calculation, we can find the moles of sodium acetate and convert it to moles of acetic acid using the same ratio:
moles of acetic acid = ratio x moles of sodium acetate
moles of acetic acid = (1/1.9953 x 10^10) x (0.100 M x (200.0 - V1)/1000)L
Now, we have two equations:
moles of acetic acid = 0.200 M x (V1/1000)L
moles of acetic acid = (1/1.9953 x 10^10) x (0.100 M x (200.0 - V1)/1000)L
By equating these two equations, we can solve for V1, the volume of acetic acid.