0.100 M solution of a weak acid, HX, is known to be 15% ionized. The weak acid has a molar mass of 72 g/mol.
1. What is Ka for the weak acid?
2. What is the pH of the buffer prepared by adding 10.0 g of the sodium salt of the acid (NaX) to 100.0 mL of 0.250 M HX
0.1M x 0.15 ionized = 0.015 concn.
...........HX --> H^+ +X^-
initial...0.1.....0.....0
change..-0.015...0.015..0.015
equil.....?.......?.......?
Ka = (H^+)(X^-)/(HX)
Substitute from the ICE chart and solve for Ka.
For #2, use the Henderson-Hasselbalch equation.
I am a little confused with the Henderson-Hasselbalch eqn on 2, #1 worked great, thanks!
You have 10.0 g of the Na salt. The acid has a molar mass of 72. We remove 1 for H and add 23 for Na for the molar mass of the NaX to be 94.
mols NaX = 10.0/94 = 0.106
mols acid = 0.1L x 0.25M = 0.025
The HH equation is
pH= pKa + log (base)/(acid)
pH = pKa from your value in #1.
mols base = 0.106
mols acid = 0.025
Crunch the numbers and solve for pH.
To answer these questions, we will need to use the concept of equilibrium and the equation for the ionization of the weak acid.
1. The degree of ionization can be given by the equation: α = √(Ka * C), where α is the percent ionization, Ka is the acid dissociation constant, and C is the initial concentration of the weak acid. Given that the percent ionization is 15% and the concentration is 0.100 M, we can use these values to find Ka.
Rearranging the equation: Ka = (α / C)^2 = (0.15 / 0.100)^2 = 2.25
Therefore, the acid dissociation constant (Ka) for the weak acid is 2.25.
2. To determine the pH of the buffer, we need to consider the Henderson-Hasselbalch equation, which is given by the equation: pH = pKa + log([A-]/[HA]), where pKa is the negative log of the acid dissociation constant Ka, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
Given that we added 10.0 g of the sodium salt of the acid (NaX), we need to convert it to moles. The molar mass of the weak acid is given as 72 g/mol, so the number of moles of NaX is 10.0 g / 72 g/mol ≈ 0.139 moles.
Now, we need to calculate the concentrations of [A-] and [HA]. Since NaX is the conjugate base, we can assume that it fully dissociates in solution, and the concentration of [A-] is equal to the number of moles of NaX divided by the volume of the solution in liters:
[A-] = 0.139 moles / 0.100 L = 1.39 M
The concentration of [HA] is given as 0.250 M.
Substituting these values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA]) = log(2.25) + log(1.39 / 0.250) ≈ log(2.25) + log(5.56) ≈ 0.352 + 0.744 = 1.096
Therefore, the pH of the buffer prepared by adding 10.0 g of the sodium salt of the acid (NaX) to 100.0 mL of 0.250 M HX is approximately 1.096.