and could you show up some steps, so I can understand? Thank you very much! PRE-

Question

and could you show up some steps, so I can understand?

Thank you very much!

PRE-LAB EXERCISE: 1. In atitration between HCI and NaOH, if the volume of the NaOH used is larger than the volume of HCI used, would this calculate to a base concentration larger than or smaller than the acid concentration? Briefly explain. 2. In a titration, 13.51 mL of NaOH(aa) are titrated using 12.85 mL of 0.1104 M HCl(aa) Calculate the molarity of NaOH(aq) (Notice that there is a (volume acid)/(volume base) ratio embedded in the calculation) 3. A 10.00 mL volume of vinegar (concentrated acetic acid solution) is diluted in a 100.0 mL volumetric flask to make a dilute acetic acid solution. The NaOH in #2 is used to titrate this dilute acetic acid, and 14.53 mL of this diluted acetic acid solution are neutralized by 12.51 mL NaOH of the molarity in #2. Calculate: a. Volume Base/Volume Acid Ratio in the titration (4 s.f.) b. Molarity of the diluted acetic acid using the ratio in a. (Ma Mb Ratio) c. Molarity of the Acetic acid in the concentrated acetic acid solution (vinegar) 100.0 mL and Vc = 10.00 mL) (Mc-Md"Vd/Vc where Vd Mass of acetic acid per liter of the concentrated acetic acid solution (vinegar) (the molar mass of acetic acid is 60.052 g/mol) (grams/L = mol/L * g/mol) d. e· The percent by mass of acetic acid in the concentrated acetic acid (vinegar) using the density of the vinegar of 1.011 g vinegar solution/mL vinegar solution. (The percent Acetic Acid in Vinegar is usually 4% to 8% by mass)

Explanation / Answer

Solved the first two problem, please post one more question to get answers to the problem since it contains lot of subparts

1) I am assuming that the molarity of both NaOH and HCl are same, since the data for the same is missing

Since the volume of NaOH is more than the volume of HCl

Hence the solution will be basic in nature

2)

M1(base)V1(base) = M2(base)V2(base)

assuming that the titration is done till the pH neutral stage

M1 * 13.51 = 12.85 * 0.1104

M1 = 0.1050M

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