Answer:
0.15 M
Explanation:
Hello!
In this case, via the Henderson-Hasselbach equation, we able to write:
[tex]pH=pKa+log(\frac{[base]}{[acid]} )[/tex]
Thus, since we are asked the concentration of the conjugate base ([base]), we proceed as follows:
[tex]3.27=3.39+log(\frac{[base]}{0.2M} )\\\\log(\frac{[base]}{0.2M} )=-0.12\\\\\frac{[base]}{0.2M}=10^{-0.12}[/tex]
[tex][base]=0.2M*0.759[/tex]
[tex][base]=0.15M[/tex]
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