Calculation of pH in Buffers Undergoing Chemical Reactions

Introduction

Beyond understanding the pH calculation of different buffer solutions, it is crucial to explore how such buffers behave under various chemical reactions, such as the addition of a strong base. In this article, we will calculate the pH of a buffer solution consisting of NH3 and NH4Cl, and then determine the pH after adding a small amount of NaOH.

Calculation of Initial pH of the Buffer Solution

The buffer solution is composed of 0.020 M NH3 and 0.030 M NH4Cl. The pKa of NH4Cl is 9.24.

Using the Henderson-Hasselbalch equation,

[ pH pK_a logdfrac{[NH_3]}{[NH_4^ ]} ]

Substitute the values:

[ pH 9.24 logdfrac{0.020}{0.030} ]

[ pH 9.24 log(0.667) ]

[ pH 9.24 - 0.176 ]

Initial pH 9.06

Effect of Adding NaOH to the Buffer Solution

Next, we consider the effect of adding 1.00 mL of 0.10 M NaOH to 0.10 L of the buffer solution. The reaction involved is:

[ OH^{-} NH_4^{ } rightarrow NH_3 H_2O ]

Calculate the moles of NaOH added:

[ moles, of, NaOH 0.10, M times 0.0010, L 0.0001, mol ]

The addition of NaOH will react with NH4Cl to form NH3.

After the reaction, the new moles of NH3 and NH4 Cl are:

[ moles, of, NH_3 0.0020 0.0001 0.0021, mol ]

[ moles, of, NH_4^ 0.0030 - 0.0001 0.0029, mol ]

Now, apply the Henderson-Hasselbalch equation again to find the new pH:

[ pH 9.24 logdfrac{0.0021}{0.0029} ]

[ pH 9.24 log(0.724) ]

[ pH 9.24 - 0.139 ]

New pH 9.10

Calculation for a Different Set of Buffer Components

For a buffer solution with Kb instead of Ka, we use the formula:

[ pOH pK_b logdfrac{[salt]}{[base]} ]

Here, let's consider the buffer with Na2CO3 and H2CO3. Given pK_b 4.75:

[ pOH 4.75 logdfrac{0.03}{0.020} ]

[ pOH 4.75 log(1.5) ]

[ pOH 4.75 0.18 ]

[ pOH 4.93 ]

Finally, calculate the pH:

[ pH 14 - pOH 14 - 4.93 9.07 ]

When 1 mL of 0.10 M NaOH is added to 1 L of the buffer, the dilution factor is 1:1000. This changes the concentration of NaOH to 0.0001 M.

Final pH Calculation After Adding NaOH

The reaction is:

[ OH^{-} NH_4^{ } rightarrow NH_3 H_2O ]

Consequently, [NH3] increases by 0.0001 M, and [NH4 ] decreases by 0.0001 M.

Now, recalculate the pOH:

[ pOH 4.75 logdfrac{0.0299}{0.02001} ]

[ pOH 4.75 log(1.5) ]

[ pOH 4.75 0.15 ]

[ pOH 4.90 ]

Converting to pH:

[ pH 14 - pOH 14 - 4.90 9.10 ]

Conclusion

The pH calculations above demonstrate the dynamic behavior of buffer solutions under the addition of strong bases or acids. Understanding these calculations is essential for the proper use of buffers in various industrial and laboratory applications.