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.
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
3 answers
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.
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.
0 answers