A buffer prepared by dissolving oxalic acid dihydrate (H2C2O4·2H2O) and disodium oxalate (Na2C2O4) in 1.00 L of water has a pH of 5.007. How many grams of oxalic acid dihydrate (MW = 126.07 g/mol) and disodium oxalate (MW = 133.99 g/mol) were required to prepare this buffer if the total oxalate concentration is 0.116 M? Oxalic acid has pKa values of 1.250 (pKa1) and 4.266 (pKa2).
To solve this problem, we need to calculate the amounts of oxalic acid dihydrate and disodium oxalate required to prepare the buffer.
Step 1: Calculate the concentrations of oxalic acid and its conjugate base in the buffer solution.
Since the buffer solution has a pH of 5.007, it is slightly acidic. Therefore, the concentration of oxalic acid (H2C2O4) would be dominant over the concentration of the conjugate base (C2O4^2-).
To calculate the concentrations, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
pKa1 = 1.250
pKa2 = 4.266
pH = 5.007
[H2C2O4] / [C2O4^2-] = 10^(pH - pKa2)
[H2C2O4] / [C2O4^2-] = 10^(5.007 - 4.266)
[H2C2O4] / [C2O4^2-] = 10^0.741
[H2C2O4] / [C2O4^2-] = 5.958
We are given that the total oxalate (C2O4^2-) concentration is 0.116 M.
Let's call the concentration of oxalic acid dihydrate (H2C2O4·2H2O) x (in moles) and the concentration of disodium oxalate (Na2C2O4) y (in moles).
Therefore, we have the following equations:
[H2C2O4] = x
[C2O4^2-] = y
[H2C2O4] + [C2O4^2-] = 0.116 M (Total oxalate concentration)
x + y = 0.116
Using the previous equation, [H2C2O4] / [C2O4^2-] = 5.958, we can substitute [H2C2O4] with 5.958 * [C2O4^2-]:
(5.958 * y) / y = 5.958
Simplifying, we get:
5.958 = 5.958
This equation confirms that [H2C2O4] / [C2O4^2-] = 5.958 holds.
Step 2: Calculate the moles of oxalic acid dihydrate.
Using the molecular weight of oxalic acid dihydrate (MW = 126.07 g/mol) and its concentration (x moles):
x = [H2C2O4] * Volume = 5.958 * y * Volume
Step 3: Calculate the moles of disodium oxalate.
Using the molecular weight of disodium oxalate (MW = 133.99 g/mol) and its concentration (y moles):
y = [C2O4^2-] * Volume = 0.116 M * Volume
Step 4: Conversion to grams.
Since we have the moles of each compound, we can multiply the moles by their respective molecular weights to get the grams required.
Grams of oxalic acid dihydrate = x * MW (g/mol)
Grams of disodium oxalate = y * MW (g/mol)
Substituting the values into the equation, we have:
Grams of oxalic acid dihydrate = (5.958 * y * Volume) * MW (g/mol)
Grams of disodium oxalate = (0.116 M * Volume) * MW (g/mol)
Given that the total volume is 1.00 L, the values for grams of each compound can be calculated using the aforementioned equations.
To solve this problem, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa and the ratio of the concentrations of the conjugate acid and base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
pH is the measured pH of the buffer solution,
pKa is the acid dissociation constant for the conjugate acid,
[A-] is the concentration of the conjugate base, and
[HA] is the concentration of the weak acid.
In this case, oxalic acid (H2C2O4) is the weak acid, and disodium oxalate (Na2C2O4) is its conjugate base.
We are given the following information:
pH = 5.007
pKa1 = 1.250
pKa2 = 4.266
Total oxalate concentration = 0.116 M
MW (oxalic acid dihydrate) = 126.07 g/mol
MW (disodium oxalate) = 133.99 g/mol
Step 1: Calculate the concentration of the acid (oxalic acid dihydrate).
Using the Henderson-Hasselbalch equation, we know that:
pH = pKa + log([A-]/[HA])
Since [A-] and [HA] are in terms of concentration, we can rewrite the equation as:
pH = pKa + log(C[A-]/C[HA])
Where C represents concentration.
For oxalic acid (HA), the given concentrations are as follows:
C[HA] = x g/mol × (1 mol/126.07 g) × (1000 mL/1 L) = x/126.07 mol/L
For oxalate (A-), we are given that the total oxalate concentration is 0.116 M, so:
C[A-] = 0.116 M
Now we can substitute these values into the equation:
pH = pKa1 + log(0.116/ (x/126.07))
5.007 = 1.250 + log(0.116/(x/126.07))
Step 2: Solve for x to find the grams of oxalic acid dihydrate.
Rearrange the equation to isolate x:
log(0.116/(x/126.07)) = 5.007 - 1.250
log(0.116/(x/126.07)) = 3.757
Now, we can convert the logarithmic equation into an exponential form:
10^(3.757) = 0.116/(x/126.07)
Simplify the equation:
10^(3.757) = (126.07/x) × 0.116
10^(3.757) × x = 126.07 × 0.116
Finally, solve for x:
x = (10^(3.757) × 126.07 × 0.116)
Step 3: Calculate the grams of disodium oxalate.
To find the grams of disodium oxalate, we need to use the concentration formula:
C = n/V
In this case, C is the concentration of disodium oxalate, n is the number of moles, and V is the volume.
The given concentration of disodium oxalate is 0.116 M, and the volume is 1.00 L.
Rearrange the equation to isolate n:
n = C × V
n = 0.116 mol/L × 1.00 L
Now, we can calculate the moles of disodium oxalate:
n = 0.116 mol
Finally, convert moles to grams using the molar mass:
grams = n × MW
grams = 0.116 mol × 133.99 g/mol
Step 4: Calculate the final answer.
Now that we know the grams of oxalic acid dihydrate (x) and disodium oxalate (grams), we can substitute these values into the final answer:
The grams of oxalic acid dihydrate required to prepare the buffer is x g.
The grams of disodium oxalate required to prepare the buffer is (0.116 mol) × (133.99 g/mol).
Calculate both values and round them to the appropriate significant figures to get the final answer.