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Wize University Chemistry Textbook > Buffers and Titrations

Buffers Problem Type 3

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Buffers Problem Type 3

3) Calculation of the pH of a buffer solution after addition of a small amount of strong base

The strong base will completely react with the acid part of the buffer

First: HA(acid part of buffer) + OH-(strong base) → A-(base part of buffer) + H2O
  • Reaction goes to completion.
  • Put the initial moles values into an ICE Table to solve for the final moles.

Next: Take the final moles value of the conjugate acid and base and plug those values into the HH equation to solve for pH:

pH=pKa + log⁡[A−][HA]pH=pKa\ +\ \log\frac{\left[A-\right]}{\left[HA\right]}



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Example: Buffers Problem Type 3

Our buffer solution is made of 0.1mol of CH3COOH and 0.2mol of NaCH3COO in 1L solution. Ka(CH3COOH)=1.8x10-5
If we added 0.02 mol of NaOH to our solution what is the pH of our solution now?

Is NaOH a base or an acid?
Base
Is it strong or weak?
Strong
Will there be complete or incomplete dissociation?
Complete

1) Write out the chemical reaction for the dissociation of NaOH. Solve for the moles of OH-.

NaOH --> Na+ + OH-
0.02 mol 0.02 mol

2) Write the chemical reaction that shows how the buffer would react with these added OH- ions:


CH3COOH + OH- --> CH3COO- + H2O
I 0.1mol 0.02mol 0.2mol C -0.02mol -0.02mol +0.02mol E 0.08mol 0 0.22mol

3) Plug in the final moles values into the Henderson Hassalbalch equation to solve for pH:

CH3COOH + H2O eqm CH3COO- + H3O+
0.08mol 0.22mol

(we are able to plug in moles since the unit moles will cancel out)

pH=pKa + log⁡[A−][HA]pH=pKa\ +\ \log\frac{\left[A-\right]}{\left[HA\right]}
pH=?, Ka=1.8x10-5, [A-]=0.22M, [HA]=0.08M
pKa=-log(Ka)
pKa=-log(1.8x10-5)
pKa=4.74

pH=4.74 + log(0.22/0.08)
pH=4.74 + log(2.75)
pH=5.2

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Wize Tip
For buffer problems you can try to see if your answer makes logical sense!
In the last problem we did, we found that the pH of the buffer before we added any strong base was 5.04 and now we just found that the pH of the buffer solution after adding a strong base rose to 5.2. It makes sense that the pH rose because a strong base has a higher pH!

Practice: Calculate the pH of a Buffer Solution


A buffer is prepared by adding 0.021 moles of HI to 250mL of 0.71M of (CH3)2NH(aq). What is the pH of this buffer? Kb((CH3)2NH(aq))=5.4x10-4 (Assume there is no change in volume)