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Wize University Chemistry Textbook > Buffers and Titrations

Buffers Problem Type 2

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Buffers Problem Type 2

2) Calculation of the pH of a buffer solution after addition of a small amount of strong acid

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


First: A-(base part of buffer) + H3O+(strong acid) → HA(acid part of buffer) + H2O
  • Reaction goes to completion
  • Plug in the initial moles values into the ICE table to find 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 2

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 75mL of 0.1M HNO3(aq), what is the pH of our buffer solution?

Is HNO3 a base or an acid?
Acid
Is it strong or weak?
Strong
Will there be complete or incomplete dissociation?
Complete

1) Write out the chemical reaction for the dissociation of HNO3. Solve for moles of H+.

HNO3 → H+ + NO3-
V=75mL
c=0.1M
n=cv
n=(0.1)(0.075)
n=7.5x10-3 moles n=7.5x10-3 moles for H+ as well since there is a 1:1 ratio

2) Write the chemical reaction that shows how the buffer would react with these added [H+] ions:

CH3COO- + H3O+ → CH3COOH + H2O
I 0.2mol 7.5x10-3 mol 0.1mol
C -7.5x10-3 mol -7.5x10-3 mol +7.5x10-3 mol
E 0.1925mol 0 0.1075 mol

3) You can plug in the final moles values of A- and HA in the Henderson Hassalbalch equation to solve for pH:

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

pKa=-log(Ka)=4.74

pH=4.74 + log(0.1925mol/0.1075mol)
pH=4.74 + log(1.79)
pH=4.99