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akhl

@akhl

Dazzler

Posted: 14 years ago

#311

Can u post here what answer u entered for #4 ?

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akhl

@akhl

Dazzler

Posted: 14 years ago

#312

I am sure either 1 or 2 in the diagram below is correct for #3.

Try the structure on the left first.

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Shree

@shree93

Dazzler

Posted: 14 years ago

#313

A voltage of 16.0 V is applied across the two ends of a Tungsten wire with a temperature coefficient of resistivity of 4.5010^-3 1/K. Initially the wire is kept at room temperature. If the wire is cooled down to freezing temperature, then the power dissipated by the wire will:

decrease
stay the same
increase

i think it will increase, but I am not 100% sure.

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akhl

@akhl

Dazzler

Posted: 14 years ago

#314

Originally posted by: shree93
A voltage of 16.0 V is applied across the two ends of a Tungsten wire with a temperature coefficient of resistivity of 4.5010^-3 1/K. Initially the wire is kept at room temperature. If the wire is cooled down to freezing temperature, then the power dissipated by the wire will:
decrease
stay the same
increase

i think it will increase, but I am not 100% sure.

The temperature is lowered. Therefore the resistance of the wire decreases.

P = V^2/R

V remains constant and R decreases. Therefore, P will increase.

You are right. It will increase.

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Sonia

@Miggi

Inactive

Posted: 14 years ago

#315

Hi,

This is a chemistry question. I did a lab in which a strong base was added in a weak acid to make a buffer solution.

from my graph@ Half Titration point pH=4.1 and volume of NaOH=6.7ml.

Mass of ascorbic acid was 0.8994g and NaOH=0.40M

The question was to choose to points equally spaced on either side of the half titration point in the buffer zone and calculate both [HA] and

[A-] for both points?Assume graduated cylinder dilivered 50ml of water in total and assume volume of acid is neglible.

One point I chose was when pH=3.7 and volume of NaOH=4.00ml.

and the other was where pH =4.5 and volume=9.00ml.

To solve: I found out the moles of ascorbic acid to be 5.06*10^-3

and added 0.0004Lof base and 0.05L of water to 0.09L

from this I found out the concentration of [HA] to be 0.056M but I don't know how to find the equilibrim concentration using ICE table.

another approach was to use formula [HA]=(moles of acid originally-moles of base added)/total volume and for [A-]= moles OH-/total vol

but it gives me a negative number here is what I did 5.06*10^-3-0.016/0.09 and it gives -0.122.

I don't know what to do please help.

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akhl

@akhl

Dazzler

Posted: 14 years ago

#316

Initial moles of the acid = 5.06 * 10^-3 = 0.00506

At the first point, moles of NaOH = 0.40 * 4.00 * 10^-3 = 0.0016

Total volume = 50 ml + 4.00 ml = 54 ml = 0.054 L

[HA] = (original moles of acid - moles of base)/total volume

[HA] = (0.00506 - 0.0016) / 0.054

[HA] = 0.0641 M

At the second point, moles of NaOH = 0.40 * 9.00 * 10^-3 = 0.0036

Total volume = 50 ml + 9.00 ml = 59 ml = 0.059 L

[HA] = (0.00506 - 0.0036)/0.059

[HA] = 0.0247 M

You wrote 0.016. It will actually be 0.0016.

Finding [A-] values:

Note that pKa = pH at half equivalence point.

So pKa = 4.1

Now you know pKa, pH and [HA]. So can you find [A-] using Henderson equation?

Edited by akhl - 14 years ago

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Sonia

@Miggi

Inactive

Posted: 14 years ago

#317

Hey, thanks for your help. I just have one more question. When finding the [H+] for the above two points, do we still assume pH=pKa and Ka = [H3O+] or would it be different as the points are before and after half equivalence point?

Edited by Miggi - 14 years ago

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akhl

@akhl

Dazzler

Posted: 14 years ago

#318

pH and [H3O+] values change but pKa and Ka values do not change.

pKa = pH at half titration point

So at all points,

pKa = 4.1

and Ka = 10^-4.1

Ka = 7.9433 * 10^-5

You already know pH values. So find [H3O+] using the formula

[H3O+] = 10^-pH

At one point, pH = 3.7

Therefore [H3O+] = 10^-3.7 = 2.00 * 10^-4

At the second point, pH = 4.5

Therefore [H3O+] = 10^-4.5 = 3.16 * 10^-5

Edited by akhl - 14 years ago

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Sonia

@Miggi

Inactive

Posted: 14 years ago

#319

Thank you so much for your help.😊

for the first question ,I got [A-] = 0.0296M is that right?

and in the last question it says to find out

Ka for those two points too. So,shouldn't the Ka be different?

I used the formula [H3O+]=Ka[HA]/[A-] and got Ka of 9.212*10^-5 for first one and 7.804*10^-5 for

2nd point. So, does that mean it's wrong or is it right?

Edited by Miggi - 14 years ago

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akhl

@akhl

Dazzler

Posted: 14 years ago

#320

You have to calculate Ka values at both points. But both should be roughly equal, otherwise something wrong in collecting the data or reading the graph.

Ignore what I wrote earlier. In this I am giving you complete answer using ICE table.

MW of ascorbic acid = 176.13

Moles of HA added = 0.8994/176.13 = 0.005106

At the first point:

Before titration:-
Total volume = 50 ml + 4.00 ml = 54 ml
[NaOH] = moles of NaOH/total volume = 0.40 M * 4.00 ml/54 ml
[NaOH] = 0.02963 M

Moles of HA = initial moles of HA - moles of NaOH = 0.005106 - 0.40 * (4.00 * 10^-3)
Moles of HA = 0.003506
[HA] = 0.003506 moles/(54 ml) = 0.003506/(54*10^-3)
[HA] = 0.0649 M

HA --------> H+ + A-
0.0649 0 0 (I)
-x +x +x (C)
0.0649-x x x (E)

x = [H+] = 10^-pH = 10^-3.7 = 2.00 * 10^-4 = 0.0002

[A-] = x + [NaOH] = 0.0002 + 0.02963 = 0.02983 M

[HA] = 0.0649 - x = 0.0649 - 0.0002 = 0.0647 M

Ka = [H+] [A-]/[HA] = 0.0002 * 0.02983/0.0647 = 9.2 * 10^-5

At the second point:

Before titration:-
Total volume = 50 ml + 9.00 ml = 59 ml
[NaOH] = moles of NaOH/total volume = 0.40 M * 9.00 ml/59 ml
[NaOH] = 0.06102 M

Moles of HA = initial moles of HA - moles of NaOH = 0.005106 - 0.40 * (9.00 * 10^-3)
Moles of HA = 0.001506
[HA] = 0.001506 moles/(59 ml) = 0.001506/(59*10^-3)
[HA] = 0.0255 M

HA ---------> H+ + A-
0.0255 0 0 (I)
-x +x +x (C)
0.0255-x x x (E)

x = 10^-pH = 10^-4.5 = 3.1623 * 10^-5 M

[A-] = x + [NaOH] = 3.1623 * 10^-5 + 0.06102 = 0.0611 M

[HA] = 0.0255 - x = 0.0255 - 3.1623 * 10^-5 = 0.02547 M
Ka = [H+] [A-]/[HA] = (3.1623 * 10^-5) (0.0611) / 0.02547 = 7.59 * 10^-5

Edited by akhl - 14 years ago