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8.0 mL of 0.10 M HC2H3O2 is titrated with 0.10 M NaOH. Perform the following calculations
Ans:
pH of the acid before any NaOH is added is
"pH = -log[H^+]\\\\pH = -log (0.10)=1"
Volume of "0.10" M "NaOH" needed to reach the equivalence point= "0.04398 L"
So the "pH" at equivalence point will be
pH = 7.00
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