I need to create a 100ml buffer with a pH of 4.00 using 0.100M Benzoic acid (pKa = 4.20)and 0.180M sodium benzoate. What volume of each material do I need to make this buffer? I think I need to use pH=pKa + log(A/HA)but am not sure if that is correct. Any help would be appreciated.
Yes, the Henderson-Hasselbalch equation is what you use.
If x = mL sodium benzoate (base)
then 100-x = mL benzoic acid(acid).
4.00 = 4.20 + log(0.18x)/[(100-x)*0.1]
and solve for x
I think that is where I keep getting stuck, with the basic math. I have 0.631 = .180x/10-0.1x and don't know where to go from there. Math was a long time ago for me :)
0.631 = (0.180x)/[(100-x)*0.100]
0.631*[(100-x)*.100] = 0.180x
multiply 0.631*0.100 first to obtain 0.0631, then
0.0631*(100-x) = 0.180x
6.31-0.0631x = 0.180x
6.31 = 0.0631+0.180x
6.31 = 0.2431x
x = 6.31/0.2431 = 25.96 mL
100-x = 74.04 mL
These should be rounded to 26.0 mL and 74.0 mL.
You should check this by multiplying 26.0x0.180 = ?? and 74.0 x 0.100 mL = ??, substitute into the HH equation and see if you arrive at a pH of 4.00.
thank you so much. I just recently went back to school and havent had chemistry in more than 4 years and math has been even longer. I really appreciate the help :)
To create a buffer with a specific pH, you can use the Henderson-Hasselbalch equation, which you correctly mentioned: pH = pKa + log(A/HA), where pH is the desired pH, pKa is the dissociation constant of the weak acid, A is the concentration of the conjugate base, and HA is the concentration of the weak acid.
In your case, you want a pH of 4.00 and are using benzoic acid (C6H5COOH) with a pKa of 4.20 and sodium benzoate (C6H5COONa). Let's calculate the concentrations of A and HA needed.
First, determine the ratio of A/HA based on the desired pH:
pH = pKa + log(A/HA)
4.00 = 4.20 + log(A/HA)
log(A/HA) = 4.00 - 4.20
log(A/HA) = -0.20
Next, calculate the actual ratio of A/HA from the concentration values:
C6H5COOH (benzoic acid) has a concentration of 0.100M.
C6H5COONa (sodium benzoate) has a concentration of 0.180M.
Since the ratio of A/HA is a negative value, we need to use the antilog to solve for the ratio. The antilog of -0.20 is approximately 0.630. This means that the ratio of [A]/[HA] is approximately 0.630.
Now, let the volume of benzoic acid (C6H5COOH) be x mL, and the volume of sodium benzoate (C6H5COONa) be y mL.
Using the ratio, we can set up the following equation:
0.630 = (0.180 * y) / (0.100 * x)
To create a 100 mL buffer, the total volume of both solutions (x + y) must equal 100 mL.
x + y = 100
You now have two equations:
0.630 = (0.180 * y) / (0.100 * x)
x + y = 100
Simplifying the first equation:
0.630 * 0.100 * x = 0.180 * y
0.063 * x = 0.180 * y
Rewriting the second equation in terms of x:
x = 100 - y
Substituting the second equation into the first:
0.063 * (100 - y) = 0.180 * y
Solving for y:
6.3 - 0.063y = 0.18y
6.3 = 0.243y
y ≈ 25.93
Now, substitute the value of y back into x + y = 100:
x + 25.93 = 100
x ≈ 74.07
Therefore, you would need approximately 74.07 mL of benzoic acid (C6H5COOH) and approximately 25.93 mL of sodium benzoate (C6H5COONa) to create a 100 mL buffer with a pH of 4.00.