From Henderson-Hasselbalch equation:
pH=pKa+logCconj.base/Cacid
pKa=-logKa=-log1.85×10-5=4.74.
While mixing of equal volumes of two solutions the resulting volume is two times higher than each one of the initial solutions, so the resulting concentrations are two times lower:
C(CH3COOH)=C(CH3COONa)=0.1/2=0.05(mol/L).
pH=4.74+log 0.05/0.05=4.74+0=4.74.
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