Answer:
[tex]pH=4.56[/tex]
Explanation:
Hello there!
In this case, given the Henderson-Hasselbach equation, it is possible for us to compute the pH by firstly computing the concentration of the acid and the conjugate base; for this purpose we assume that the volume of the total solution is 0.025 L and the molar mass of the sodium base is 234 - 1 + 23 = 256 g/mol as one H is replaced by the Na:
[tex]n_{acid}=\frac{0.2g}{234g/mol}=0.000855mol\\\\n_{base}= \frac{0.2g}{256g/mol}=0.000781mol[/tex]
And the concentrations are:
[tex][acid]=0.000855mol/0.025L=0.0342M[/tex]
[tex][base]=0.000781mol/0.025L=0.0312M[/tex]
Then, considering that the Ka of this acid is 2.5x10⁻⁵, we obtain for the pH:
[tex]pH=-log(2.5x10^{-5})+log(\frac{0.0312M}{0.0342M} )\\\\pH=4.60-0.04\\\\pH=4.56[/tex]
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