How to use the Henderson Hassalbalch Equation

Moderators: Chem_Mod, Chem_Admin

Alex Nguyen 3I
Posts: 100
Joined: Fri Sep 25, 2015 3:00 am

How to use the Henderson Hassalbalch Equation

Postby Alex Nguyen 3I » Sat Dec 05, 2015 8:39 pm

How do you use the equation for a titration before the stoichiometric point? For example finding the pH of .100M HCOOH 25mL titrated with 5 mL of .150M NaOH. The pH turns out to be 3.38, and I tried finding the concentration using the new volumes, but it didn't turn out right.

Isabel Strouse 1H
Posts: 24
Joined: Fri Sep 25, 2015 3:00 am

Re: How to use the Henderson Hassalbalch Equation

Postby Isabel Strouse 1H » Sat Dec 05, 2015 10:07 pm

Hi Alex,

I worked out the problem, step-by-step in 10 steps. I'll explain the steps here. I hope this helps!

Step 1: Since you're given the volume of both the weak acid and the strong base you're titrating with, you can calculate the moles of both:
0.100 M HCOOH x 0.025 L = 0.0025 mol HCOOH.
0.150 M NaOH x 0.005 L = 0.00075 mol NaOH.


Step 2: Subtract the smallest number of moles from the larger, so the mol NaOH from the mol HCOOH. In essence, this acts kind of like the "limiting reactant". This also indicates that there will be no mol NaOH left over.
0.0025 mol HCOOH - 0.00075 mol NaOH = 0.00175 mol HCOOH left over.

Step 3: The equation for this reaction is HCOOH + NaOH <---> HCOO- + Na+ + H2O. When we subtracted the .00075 mol NaOH, we also have to add .00075 mol to the right side of the equation because the reaction is at equilibrium. Since Na+ is just a spectator ion and doesn't react, and H2O does not change the concentration either, we only add the moles to the HCOO-.
mol HCOO- = 0.00075 mol.

After this, we find the new concentrations of HCOOH and HCOO- using the moles we found and the total number of liters in solution, .025+.005 = .03L.
[HCOOH] = 0.00175 mol/0.03 L = 0.0583 M HCOOH.
[HCOO-] = 0.00075 mol/0.03 L = 0.025 M HCOO-.


Step 4: We can now set up an ice table for the reaction at this point. The HCOO- reacts with water to yield HCOOH and OH-.
HCOO- + H2O <---> HCOOH + OH-
I 0.025 M --- 0.0583 M 0 M
C -x --- +x +x
E 0.025-x --- 0.0583+x x


Step 5: We are given Ka for HCOOH (1.8x10^-4), but the equilibrium equation calls for a Kb. So we find Kb using KaKb = Kw.
Kb = Kw/Ka = 1.0x10^-14/1.8x10^-4 = 5.56 x 10^-11

Step 6: Set up the equation for the basicity constant and substitute in the equilibrium values of the reactants and products.
Kb = [HCOOH][OH-]/[HCOO-] = (x)(0.583+x)/(0.025-x)

Step 7: Since Kb << 10^-3, we can assume that the change in concentration is negligible and therefore the value of x will have negligible effect.
(x)(0.583+x)/(0.025-x) = 0.583(x)/0.025

Step 8: Set the equation equal to the basicity constant we found, 5.56 x 10^-11. Solve for x. X is also equal to [OH-] at equilibrium.
0.0583x/0.025 = 5.56x10^-11
so x = 2.38x10^-11 = [OH-]


Step 9: Using the equilibrium concentration of [OH-] we found, solve for the pOH.
pOH = -log[OH-] = -log(2.83x10^-11) = 10.62

Step 10: Solve for pH using pH + pOH = 14.00.
14.00-10.62=3.38

Ta da!

An 11th step we really should do is calculate the 5% rule to check to make sure our x really doesn't have an effect on the concentrations to prove our assumption is viable, but that step is pretty self explanatory. Just follow the formula!

Happy studying! :)


Return to “*Titrations & Titration Calculations”

Who is online

Users browsing this forum: No registered users and 1 guest