Calculate the KA of formic acid if 2.3 g of it is dissolved in 1 litre of water to give [H+] of 0.003 M.
mols HCOOH = grams/molar mass, then
M = mols/L soln. I get about 0.05M but you should confirm that.
.........HCOOH + H2O ==> H3O^+ + HCOO^-
I .......0.05..............0.......0
C.......-0.003.........0.003..0.003
E...0.05-0.003.........0.003...0.003
Substitute from the ICE chart above into the IKa expression and solve for Ka.
To calculate the Ka (acid dissociation constant) of formic acid, we need to use the given information and the mathematical relationship between the concentration of the ionized form of acid ([H+]), the initial concentration of the acid (C), and the equilibrium concentration of the ionized form of acid in moles per liter (x).
The balanced equation for the dissociation of formic acid (HCOOH) in water is as follows:
HCOOH ⇌ H+ + COO-
Given:
- Mass of formic acid (HCOOH) = 2.3 g
- Volume of water = 1 L
- Concentration of [H+] = 0.003 M
First, we need to convert the mass of formic acid (HCOOH) to moles. To do this, we will use the molar mass of formic acid:
Molar mass of HCOOH = (1 * atomic mass of H) + (1 * atomic mass of C) + (2 * atomic mass of O)
= (1 * 1.008 g/mol) + (1 * 12.01 g/mol) + (2 * 16.00 g/mol)
= 46.03 g/mol
Now, we can calculate the moles of formic acid (HCOOH):
moles of HCOOH = mass of HCOOH / molar mass of HCOOH
= 2.3 g / 46.03 g/mol
= 0.0499 mol
The concentration of formic acid (HCOOH) in moles per liter is the moles of HCOOH divided by the volume of water:
C = moles of HCOOH / volume of water
= 0.0499 mol / 1 L
= 0.0499 M
Since formic acid (HCOOH) dissociates into one H+ ion, the concentration of [H+] is equal to x.
[H+] = x = 0.003 M
Now, we can set up the equation for Ka:
Ka = [H+] * [COO-] / [HCOOH]
Since the concentration of [H+] is given, and the concentration of [COO-] is equal to x, we have:
Ka = 0.003 * x / 0.0499
Simplifying, we get:
Ka = 0.06012 * x
To calculate the value of Ka, we need to determine the value of x. We can use the quadratic formula since the equation is quadratic. Rearranging the equation, we get:
0.06012 * x - Ka = 0
Using the quadratic formula, where a = 0.06012, b = -Ka, and c = 0:
x = (-b ± √(b^2 - 4ac)) / 2a
Since we are given the concentration of [H+] and formic acid does not ionize completely, the value of x will be small compared to the concentration of formic acid. Therefore, we can assume that x is much smaller than 0.0499.
Using this assumption, we can approximate that [H+] ≈ x. Therefore, we can substitute [H+] with x in the equation:
Ka = 0.06012 * x
Now, we substitute the value of [H+] into the equation:
Ka = 0.06012 * 0.003
= 0.00018036
Therefore, the Ka value of formic acid is approximately 0.00018036 (rounded to the appropriate number of significant figures).
Note: This approximation is valid because the value of x is much smaller than the initial concentration of formic acid (0.0499 M). However, if a higher level of precision is required, the quadratic formula should be used to solve for x, and then the value of Ka can be calculated accordingly.