How many grams of formic acid, HCOOH, must be added to 0.270L of 0.180M sodium formate, HCOONa, to yield a buffer solution with a pH of 4.20? Assume no volume change.
Ka of HCOOH= 5.010^-4
The Henderson-Hasselbalch equation.
To calculate the amount of formic acid (HCOOH) needed to make a buffer solution with a pH of 4.20, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this case, formic acid (HA) is the acidic component, and sodium formate (HCOONa) is the conjugate base (A-). We want to achieve a pH of 4.20 and the pKa value for formic acid is given as 5.010^-4.
Rearranging the Henderson-Hasselbalch equation, we get:
log([A-]/[HA]) = pH - pKa
Substituting in the given values, we have:
log([A-]/[HA]) = 4.20 - (-4.3010)
log([A-]/[HA]) = 8.5010
Now, let's determine the ratio of [A-]/[HA] using the antilog function:
[A-]/[HA] = antilog(8.5010)
[A-]/[HA] = 10^(8.5010)
[A-]/[HA] = 31810.884
This means that the ratio of [A-]/[HA] should be 31810.884.
Since we assume no volume change and the initial volume is given as 0.270L, the moles of sodium formate can be calculated as follows:
moles of sodium formate = concentration of sodium formate * volume
moles of sodium formate = 0.180M * 0.270L
moles of sodium formate = 0.0486 mol
As we have established that the ratio of [A-]/[HA] should be 31810.884, we can set up the following equation using stoichiometry and the balanced chemical equation:
31810.884 = moles of sodium formate / moles of formic acid
Solving for moles of formic acid:
moles of formic acid = moles of sodium formate / 31810.884
moles of formic acid = 0.0486 mol / 31810.884
moles of formic acid ≈ 1.53 * 10^-6 mol
Finally, let's convert moles of formic acid to grams using its molar mass:
grams of formic acid = moles of formic acid * molar mass of formic acid
The molar mass of formic acid (HCOOH) is:
(1 mol of H = 1.00784 g/mol, 1 mol of C = 12.0107 g/mol, and 2 mol of O = 15.9994 g/mol)
molar mass of formic acid = 1.00784 + 12.0107 + 15.9994 + 15.9994 = 46.00734 g/mol
grams of formic acid = (1.53 * 10^-6 mol) * (46.00734 g/mol)
grams of formic acid ≈ 7.04 * 10^-5 g
Therefore, approximately 7.04 * 10^-5 grams of formic acid (HCOOH) must be added to 0.270L of 0.180M sodium formate (HCOONa) to yield a buffer solution with a pH of 4.20.
To find the amount of formic acid (HCOOH) needed to make a buffer solution with a pH of 4.20, we need to apply the Henderson-Hasselbalch equation for buffer solutions.
The Henderson-Hasselbalch equation is given as:
pH = pKa + log ([salt]/[acid])
In this case, the acid is formic acid (HCOOH), and the salt is sodium formate (HCOONa). We are given the pH (4.20) and the Ka value for formic acid (5.010^-4). The concentration of sodium formate is also given (0.180 M).
First, let's rearrange the equation and solve for [acid]:
pH - pKa = log ([salt]/[acid])
10^(pH - pKa) = [salt]/[acid]
[acid] = [salt] / 10^(pH - pKa)
Now, let's calculate the concentration of formic acid:
[acid] = 0.180 M / 10^(4.20 - (-log10(5.010^-4)))
To perform this calculation, we need to evaluate 10^(4.20 - (-log10(5.010^-4))).
1. Start by calculating log10(5.010^-4):
log10(5.010^-4) ≈ -3.30
2. Then, calculate 4.20 + 3.30:
4.20 + (-3.30) ≈ 0.90
3. Finally, evaluate 10^0.90:
10^0.90 ≈ 7.943
Now, let's substitute this value back into the equation to find the concentration of formic acid:
[acid] = 0.180 M / 7.943
[acid] ≈ 0.0226 M
Lastly, we need to calculate the number of grams of formic acid needed. Since we have the volume (0.270 L) and the concentration (0.0226 M), we can use the formula:
grams = concentration (M) x volume (L) x molar mass (g/mol)
The molar mass of formic acid (HCOOH) is approximately 46.03 g/mol.
grams = 0.0226 M x 0.270 L x 46.03 g/mol
grams ≈ 0.335 g
Therefore, approximately 0.335 grams of formic acid (HCOOH) would need to be added to 0.270L of 0.180M sodium formate (HCOONa) to yield a buffer solution with a pH of 4.20.