How many grams of potassium formate, KHCOO, must be added to 500 mL of a .07 M solution of HCOOH to produce a buffer solution with a pH of 3.50?
given: Ka of HCOOH=1.8 x 10^-4
Use the Henderson-Hasselbalch equation and solve for (base)/(acid) concns. Calculate moles acid in the problem, substitute into the B/A ratio and solve for base concn. Then g = moles x molar mass for HCOOK grams.
To determine the amount of potassium formate (KHCOO) needed to create a buffer solution with a pH of 3.50, we first need to understand the components and principles of a buffer solution.
A buffer solution is composed of a weak acid and its conjugate base (or a weak base and its conjugate acid). In this case, the weak acid is formic acid (HCOOH), and its conjugate base is potassium formate (KHCOO).
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the concentration of its acid and conjugate base:
pH = pKa + log [A-]/[HA]
Where:
pH = desired pH of the buffer solution
pKa = -log10 (Ka), which is the logarithmic expression of the acid dissociation constant of the weak acid
[A-] = concentration of the conjugate base
[HA] = concentration of the weak acid
In this case, we are given the pKa value of formic acid (HCOOH) as 1.8 x 10^-4 and the desired pH as 3.50.
First, we need to find the concentration of the weak acid [HA] required to achieve the desired pH. We can rearrange the Henderson-Hasselbalch equation and solve for [HA]:
pH - pKa = log [A-]/[HA]
10^(pH - pKa) = [A-]/[HA] (converting back from logarithmic form)
10^(3.50 - (-log10(pKa))) = [A-]/[HA]
10^3.50/(1.8 x 10^-4) = [A-]/[HA]
1967.65/(1.8 x 10^-4) = [A-]/[HA]
10931.39 = [A-]/[HA]
We know that [A-] + [HA] = total concentration of HCOOH (0.07 M, as given). So, we can write:
[A-] + [HA] = 0.07
Substituting the value of [A-] from the equation above, we get:
10931.39 + [HA] = 0.07
[HA] = 0.07 - 10931.39
[HA] = -10931.32 M
However, since the concentration of an acid cannot be negative, we can conclude that the concentration of [HA] is essentially zero.
Now, we need to find the concentration of the conjugate base, [A-]:
[A-] = 10931.39 M
Finally, we can use the formula for concentration:
Concentration (in grams/L) = (Molarity x Molecular Weight) / 1000
The molecular weight of potassium formate (KHCOO) is 98.11 g/mol.
Concentration (in grams/L) = (10931.39 x 98.11) / 1000
Concentration (in grams/L) = 1071.36 g/L
To find the mass of potassium formate needed to make 500 mL of the buffer solution, we use the following formula:
Mass (in grams) = Concentration (in grams/L) x Volume (in L)
Mass (in grams) = 1071.36 g/L x (500 mL / 1000)
Mass (in grams) = 535.68 g
Therefore, approximately 535.68 grams of potassium formate (KHCOO) must be added to 500 mL of a 0.07 M solution of HCOOH to produce a buffer solution with a pH of 3.50.