# Buffer solutions

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## Example 3

An acetic acid / acetate buffer with excess acid [HAc] = 1.6 $molL.-1$ and [Ac] = 0,4 $molL.-1$ should as above with 0.1 $molL.-1$ NaOH are added. Of the $pH$-Value of the buffer is initially calculated according to:

$pH=pKa+lg([Ac-][HAc])=4,75+lg(0,41,6)=4,15$

After adding NaOH the following results:

$pH=pKa+lg([Ac-]+[NaOH][HAc]−[NaOH])=4,75+lg(0,4+0,11,6−0,1)=4,25⇒ΔpH=0,12$

the $pH$-The change is more pronounced than with the 1: 1 buffer, so the buffer capacity is lower.