Answer to Question #96881 in Inorganic Chemistry for Muhammed bashiru

Question #96881
if 24.83cm3 of modm-3 NAOH is titrated to it's end point with 39.45cm3 of HCL, what is the molar concentration of the HCL.
1
Expert's answer
2019-10-21T07:16:57-0400

Volume of "NaOH(V_1)=24.83 \\ cm^3"

Concentration of "NaOH(M_1)=mol\/dm^3=10^{-3} mol\/cm^3"

Volume of "HCl(V_2)=39.45 \\ cm^3"

Using molarity equation,"M_1V_1=M_2V_2"

"10^{-3}\\times 24.83=39.45\\times M_2"

"M_2=0.6294\\times 10^{-3} mol\/cm^3=0.6294 mol\/dm^{3}=0.6294 mol\/L"


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