Question

Experiment 10 An Acid-Base Neutralization Reaction A. Determination of Unknown Acid Concentration Trial Trial 2 Trial 3 Trial 4 Volume of acid 10m 10m 10m 10mL Final buret reading- NaOH (mL) To.8ML 23.8ml 26.2m4 29.4m Initial buret reading- NaOH (mL) Volume of NaOH required to reach the end point. V 10.8m. 13ML 2.4mL 3.2mL Molarity of the acid, M. 10. OmL/108m./23.8mL 26.2mL) Unknown acid number: 16 Molarity of NaOH: Average Molarity of acid 0-1076m Show all calculations for trials 1 to 4 below:

Answer 1

The acid base neutralization reaction is given by M1V1 (acid) = M2V2 (base). The values of volumes of acid and base are given and also the value of molraity of acid is given. So, by substituting the values of volume of base for different trials, the molarity of the base is calculated and all of them can be averaged. The detailed solution is given below:

Answer: Average molarity of acid = 0.1076 M . Volume of aid = 10 mL Triall MV : M₂ V, - is the Neutralization equations . where Mi u the Molarity of acid, and - Vi is the volume of acid. - Mz and 2 are Molarity of Base and Voliese of Body respectively, son 0.1076 MX IOML = M, XC10.8 mL). Merose 0.0996 M Trial 2 0.1076 x 10 = MX 13 IM Base = 0.0827M )

Irial-3 _ 0.1076 X10 = MX 2.4 TM Base 3 0.448 MT - Trial-4 0.1076 810= MX 3.2 Mease = 0.336M