Business > CASE STUDY > module5 Lab2.docx CHM1100C Module 5 Lab 2 Rasmussen University CHM1100C Section 08 Gene (All)
module5 Lab2.docx CHM1100C Module 5 Lab 2 Rasmussen University CHM1100C Section 08 Gene
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module5 Lab2.docx CHM1100C Module 5 Lab 2 Rasmussen University CHM1100C Section 08 General Chemistry Sabina Lab 2 120 ml of water is put in 80 ml of 10.0M NaOH, so molarity of NaOH solution 2.6 ... 2 mLˆ—10 M 30 mL =0.873 M Now, on titrating this NaOH solution with 100 mL of the acid solution, the volume of NaOH required is 39.3 mL, So, volume of NaOH ˆ—concentration of NaOH =volume of acidˆ—concentration of acid So, 39.3 mLˆ—0.873 M =100 mLˆ—concentration of acid So, concentration of the acid =2.915 M 1. 3labquestionsdealingwiththeAcidLab(below) 1. Briefly describe the equivalence point in a titration. What species are present? The equivalence point in a titration is the point where amount of +¿ ions are the same as the amount of ˆ’¿ ions, so if both acid and base are monoprotic, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . [Show More]
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