Determination of the pH Scale by the Method of Successive Dilutions

Anonymous
timer Asked: May 28th, 2019
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Question Description

This is an online virtual lab. The link will be below. All you have to do is the work that is UNDERLINED. As you do the virtual lab online, PLEASE TAKE SCREENSHOTS OF THE RESULTS of each experiment. And then answer that one question that is underlined below. Please do your best work, this is my final. Thank you!

http://chemcollective.org/vlab/100

Objective:

It is fairly common knowledge that neutral water has a pH of 7, acids have a pH <7 and bases have a pH>7, but few people understand this in terms of the actual hydronium ion concentration. Our objective is to develop an understanding of logarithmic scales by developing a pH scale

Background:

The pH scale describes the hydronium ion concentration in aqueous systems

pH = -log[H3O+]

[H3O+] = 10-pH = 1/10pH


The Method of Successive Dilutions
is an experimental technique for preparing a series of solutions of different concentrations from one volume of stock solution.

Lets look at a series of half dilutions.

With the virtual lab fill 5 flasks with a constant amount of water (less than half the volume of the flask), for simplicity, we will use 20 ml, but any amount will do.

*Virtual Lab Tip* - Right Click on each flask and label it

Now add the same amount of stock 1M HCl to the first flask (20 mL), and note that the concentration has been diluted in half, [H3O+] = 0.500M or 1/2 (1/21) the original molarity.

From this flask transfer 20 mL to the second flask and note the it has been diluted in half again,[H3O+]= 0.250M or is one fourth (1/22) the concentration of the original stock solution. ( one screenshot here)

Repeating this procedure with the remaining 3 flasks gives:

3rd dilution: [H3O+] = 0.12500 or 1/8 (1/23) the original stock molarity (one screenshot here)

4th dilution: [H3O+] = 0.06250 or 1/16 (1/24) the original stock molarity (one screenshot here)

5th dilution: [H3O+] = 0.03125 or 1/32 (1/25) the original stock solution. (one screenshot here)

Lets look at this in more detail:

[H3O+] = 2-n = 1/2n

Where n is the number of successive dilutions and by using a dilution factor of one to two, you have come up with a log base 2 scale.

Question: Would changing the volume of the original stock solution and the incremental dilution volumes to a new constant value effect the successive concentrations. Say by starting with 10 mL and transferring 10 mL increments? If you say yes, repeat the above with 10 mL increments and explain. ANSWER THIS QUESTION IN COMPLETE SENTENCES. NO PLAGIARISM. IF YOU DO ANY EXPERIMENTS FOR THIS QUESTION, PLEASE SCREENSHOT THE RESULTS.

Tutor Answer

jlco88
School: Duke University

Hey!. Here you go. I added every screenshot as requested and the final answer. Please note I didn´t include the question itself in the document as it checks out for plagiarism. Please tell me if you have any doubts.

Screenshot 1. Second dilution

Screenshot 2. Third dilution

Screenshot 3. Fourth dilution

Screen...

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Review

Anonymous
awesome work thanks

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