Question

An unknown Acid was titrated three times with the same standardized base. The volume of base for each titration was 15.56 mL 15.52 mL and 16.05 mL. What is the average amound of base used to titrate the unkown Acid?

Answer 1

Answer: (15.56mL +15.52mL)/2=15.54mL

`Average:(15.56mL+15.52mL)/(2)=(31.08mL)/(2)=15.54mL`

Explanation: The volume reading is called a titer value. For precision the values should be within 0.05mL of each other. Values that do not fall within this region is called an outlier is not used to find the average titer value.