8.0 mL of 0.10 M NH3 is titrated with 0.10 M HCl. Perform the following calculations (15 points):
pH of the NH3 before any HCl is added 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= 0.04398L0.04398 L0.04398L
PH=4.3P_H = 4.3PH=4.3
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