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A 25.0 mL solution of 0.817 mol L-1 NH3 is titrated using 0.669 mol L-1 HCl....

A 25.0 mL solution of 0.817 mol L-1 NH3 is titrated using 0.669 mol L-1 HCl. What volume of NaOH (in mL) is needed to reach the equivalence point in this experiment?

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Answer #1

find the volume of HCl used to reach equivalence point

M(NH3)*V(NH3) =M(HCl)*V(HCl)

0.817 M *25.0 mL = 0.669 M *V(HCl)

V(HCl) = 30.5 mL

Answer: 30.5 mL

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