You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.220 M sodium benzoate.
How much of each solution should be mixed to prepare this buffer?
mL of benzoic acid ?
mL of sodium benzoate?
To prepare a buffer solution, you need to mix a weak acid and its conjugate base. In this case, the weak acid is benzoic acid (C6H5COOH) and its conjugate base is sodium benzoate (C6H5COONa).
To calculate the volumes of the solutions required, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([conjugate base]/[weak acid])
Given:
pH = 4.00
pKa = 4.20
Concentration of benzoic acid ([weak acid]) = 0.100 M
Concentration of sodium benzoate ([conjugate base]) = 0.220 M
Let's substitute the given values into the Henderson-Hasselbalch equation:
4.00 = 4.20 + log([0.220]/[0.100])
Next, we can rearrange the equation to solve for the ratio [conjugate base]/[weak acid]:
log([0.220]/[0.100]) = 4.00 - 4.20
log([0.220]/[0.100]) = -0.20
Now, take the antilog of both sides of the equation to solve for the ratio [conjugate base]/[weak acid]:
[0.220]/[0.100] = 10^(-0.20)
Simplifying the right side:
[0.220]/[0.100] = 0.631
Therefore, the ratio [conjugate base]/[weak acid] is approximately 0.631.
Now, let's find the volumes of each solution required to make a total volume of 100.0 mL.
Let x be the volume (in mL) of benzoic acid.
Then, (100.0 - x) will be the volume (in mL) of sodium benzoate.
We can set up an equation using the ratio of concentrations:
0.220/0.100 = (100.0 - x)/x
Now, solve for x:
0.220x = 0.100(100.0 - x)
0.220x = 10 - 0.100x
0.320x = 10
x ≈ 31.25 mL (rounded to 2 decimal places)
Therefore, you would need approximately 31.25 mL of benzoic acid and approximately (100.0 - 31.25) = 68.75 mL of sodium benzoate to prepare the pH=4.00 buffer solution.
pH = pKa + log (base)/(acid)
Let x = mL base
then 100-x = mL acid
millimols base = mL x M
millimols acid = mL x M
Substitute into the HH equation and solve for x = mlL base and 100-x = mL acid. That's it.