Set up two data tablet that are appropriate for this experiment. Do not turn the

Question

Set up two data tablet that are appropriate for this experiment. Do not turn these in as they will be evaluated instructor during the laboratory period. Refer to your textbook at some titration curves specifically looking at the difference between strong base titration and a weak acid/strong base titration. cate the directions for using a pH meter in the back of your lab manual and skim them. You will not be allowed to begin the lab unless the following are correctly completed. Calculate the amount of solid sodium hydroxide and the amount of water needed to prepare the 0.1 M NaOH solution that will be used in the titration. Calculate the volume of 6M sodium hydroxide and the amount of water needed to prepare the 0.1 M NaOH solution that will be used in the titration. Assume that volumes are additive. If 25.00 ml of 0.10 M acetic acid is titrated with 0.10 M NaOH, what volume of NaOH will it take to reach equivalence point? If 25 00 ml of 0 10 M hydrochloric acid is titrated with 010 M NaOH, what volume of NaOH will it take to the following questions are based on the titration part of Experiment 1-Graphing Laboratory Data. Use data, not the graph, to answer these questions Between what two volumes, that were read on the burette, did the pH change the most? How much base was added for this pH change and what was the pH change? One milliliter of an aqueous solution is usually considered to be equal to 20 drops of that solution. Is it possible to dispense 0.1 ml of an aqueous solution from a burette given that a burette can dispense quarter d Explain. Write the Lewis structure for CH_3COOH. Include lone pairs and all bonds. Circle the acidic hydrogen atom.

Explanation / Answer

Q1.

solid NaOH required for:

1 M solution

Assume we will feel a burte, so we need 50 mL of base

so

V = 50 mL = 50*10^-3 L

mol = MV = 1*50*!0^-3 = 50*10^-3 moles of NaOH

mass = mol*MW = (50*10^-3)(39.997) = 1.99985 g of NaOH required

Q2.

from 6 M

so..

50 mL required...

M1V1 = M2V2

6*V1 = 0.1*50

V1 = 0.1*50/6 = 0.83333 mL of stock solution

Q3.

for equivalence point

mmol of acid = MV= 25*0.1= 2.5 mmol

so mmol of base reuqired = 2.5

so

M = mmol/mL

mL = mmol/M = 2.5/0.1 = 25 mL required

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