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8.0 mL of 0.10 M NH3 is titrated with 0.10 M HCl. Perform the following calculations
pH of the base before any HCl is addedi is
PH=−log(0.10)=1PH = -log (0.10)=1PH=−log(0.10)=1
Volume of 0.10 M HCl needed to reach the equivalence point= 8.0L8.0 L8.0L
pH = 7.00
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