Solution.
"V(HI) = 120.0mL=0.120L;"
"M(HI) =0.50M;"
"V(HClO_4) =850mL=0.850L;"
"M(HClO_4)=0.05M;"
"\\nu(HI)=V(HI)\\sdot M(HI); \\nu(HI)=0.120L\\sdot0.50M=0.06mol;"
"\\nu(HClO_4)=V(HClO_4)\\sdot M(HClO_4);"
"\\nu(HClO_4)=0.850L\\sdot0.05M=0.0425mol;"
"\\nu_{final}=0.06mol+0.0425mol=0.1025mol;"
"[H_3O^+]=\\dfrac{\\nu_{final}}{V_{final}}; [H_3O^+]=\\dfrac{0.1025mol}{0.970L}=0.1057M;"
"pH=-lg[H_3O^+]=-lg(0.1057)=lg9.46=0.976" ;
Answer: "pH=0.976."
Comments
Leave a comment