How many millilitres of 0.0890 mol L^-1 NaOH are required to titrate 25.0 mL of 0.0680 $\operatorname { mol } \mathrm { L } ^ { - 1 } \mathrm { HBr }$ to the equivalence point? A)3.49 B)0.151 C)32.7 D)0.242 E)19.1

Question

Quizplus · Accepted Answer

The moles of HBr are 0.0680 mol/L * 0.025 L = 0.0017 mol. Since the reaction between NaOH and HBr is 1:1, the moles of NaOH needed are also 0.0017 mol. The volume of NaOH required is 0.0017 mol / 0.0890 mol/L = 0.0191 L, which is 19.1 mL.

How Many Millilitres of 0 mol⁡L−1HBr\operatorname { mol } \mathrm { L } ^ { - 1 } \mathrm { HBr }molL−1HBr

How Many Millilitres of 0 $\operatorname { mol } \mathrm { L } ^ { - 1 } \mathrm { HBr }$