You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.240 M sodium benzoate.
How much of each solution should be mixed to prepare this buffer?
pH = pKa + log(b/a)
Substitute and I get
b/a= 0.631 so b=0.631*a or
1.585*a = b
Let x = mL of the base(b) and 100-x = mL acid(a)
Then 1.585*mols b = mols a.
1.585(0.240x) = (100-x)*0.1
Solve for x = mL base and 100-x = mL acid.
I get approximately 20 mL base and approximately 80 mL acid but you need to go through and get a better set of numbers. I ALWAYS check it.You should, too.
mmols base = 20 x 0.240 =about 4.8
mmols acid = 80 x 0.1 = about 8.0
pH = 4.2 + log(4.8/8.0) = 4.2 + (-.22) = about 3.98. I think if you put in the correct numbers it will work ok to give 4.00.
To prepare the pH=4.00 buffer solution using benzoic acid and sodium benzoate, we can follow these steps:
Step 1: Calculate the mole ratio between benzoic acid and sodium benzoate:
Since the pKa of benzoic acid is 4.20 and the desired pH is 4.00, we can assume that the concentration of benzoic acid and benzoate ion will be approximately equal in the buffer solution.
Step 2: Calculate the concentration of benzoic acid needed:
pKa = -log(Ka)
4.20 = -log(Ka)
Ka = 10^(-4.20)
Ka = 6.31 x 10^(-5)
[H+][A-] / [HA] = Ka
[x][x] / [0.100 - x] = 6.31 x 10^(-5)
x^2 / (0.100 - x) = 6.31 x 10^(-5)
Since x is much smaller than 0.100, we can assume that (0.100 - x) ≈ 0.100
x^2 / 0.100 = 6.31 x 10^(-5)
x^2 = 6.31 x 10^(-5) * 0.100
x^2 = 6.31 x 10^(-6)
x ≈ 7.94 x 10^(-3)
[HA] = 0.100 - 7.94 x 10^(-3)
[HA] ≈ 0.0921 M
Step 3: Calculate the moles of benzoic acid needed:
moles of benzoic acid = [HA] * volume (L)
moles of benzoic acid = 0.0921 mol/L * 0.100 L
moles of benzoic acid = 0.00921 mol
Step 4: Calculate the volume of 0.100 M benzoic acid solution needed:
Moles = Concentration * Volume
0.00921 mol = 0.100 mol/L * volume (L)
volume (L) = 0.00921 mol / 0.100 mol/L
volume (L) = 0.0921 L
Step 5: Calculate the volume of 0.240 M sodium benzoate solution needed:
Since we want the concentration of benzoate ion to be approximately equal to benzoic acid, we will use the same volume calculated in step 4, 0.0921 L.
Therefore, to prepare the pH=4.00 buffer solution, you should mix 0.0921 L (or 92.1 mL) of 0.100 M benzoic acid solution with 0.0921 L (or 92.1 mL) of 0.240 M sodium benzoate solution.
To calculate how much of each solution should be mixed, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where [A-] is the concentration of the conjugate base (sodium benzoate) and [HA] is the concentration of the acid (benzoic acid).
In this case, we have:
pH = 4.00
pKa = 4.20
Let's solve the equation for [A-]/[HA]:
4.00 = 4.20 + log([A-]/[HA])
Rearranging the equation:
log([A-]/[HA]) = 4.00 - 4.20
log([A-]/[HA]) = -0.20
Now we can take the antilog of both sides to get rid of the logarithm:
[A-]/[HA] = 10^(-0.20)
[A-]/[HA] = 0.63
This means that the concentration of sodium benzoate ([A-]) should be 0.63 times the concentration of benzoic acid ([HA]).
Next, we need to determine the total moles of benzoic acid and sodium benzoate required to make 100.0 mL of the buffer solution.
Volume of solution = 100.0 mL = 0.100 L
Let x be the moles of benzoic acid needed, and let y be the moles of sodium benzoate needed.
Using the given concentrations:
x = 0.100 M * 0.100 L = 0.010 moles
We know that [A-]/[HA] = 0.63, so:
y = 0.63 * x = 0.63 * 0.010 = 0.0063 moles
Now, we can calculate the volume of each solution needed.
Volume of benzoic acid solution = moles / concentration = 0.010 moles / 0.100 M = 0.100 L = 100.0 mL
Volume of sodium benzoate solution = moles / concentration = 0.0063 moles / 0.240 M = 0.026 L = 26.0 mL
To prepare the buffer solution, mix 100.0 mL of the 0.100 M benzoic acid solution with 26.0 mL of the 0.240 M sodium benzoate solution.