G.C. Problem

[dwcardman]

Opening myself up for some abuse here, but what the heck:

(From Topscore)

What volume of HCl was added if 20mL of 1.0M NaOH was titrated with 1.0M HCl to produce a pH=2?

a) 10.2 mL
b) 20.0 mL
c) 30.4 mL
d) 35.5 mL
e) none of these

ANSWER: e

It's obviously over 20 mL, but how far? What am I missing?

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[thehipster]

Already been discussed in the following thread:

http://forums.studentdoctor.net/showthread.php?t=299979

 

[dat_student]

Opening myself up for some abuse here, but what the heck:

(From Topscore)

What volume of HCl was added if 20mL of 1.0M NaOH was titrated with 1.0M HCl to produce a pH=2?

a) 10.2 mL
b) 20.0 mL
c) 30.4 mL
d) 35.5 mL
e) none of these

ANSWER: e

It's obviously over 20 mL, but how far? What am I missing?

V = Volume
pH = 2
-log [H+] = 2
[H+] = 10 ^ -2

(V * 1.0 M HCl) - (20 * 1.0 M OH-)
---------------------------------- = 10^-2
V + 20

V - 20
------- = 10 ^ -2
V + 20

0.99 V = 20.2

V = ~20.4 mL

 

[allstardentist]

its 20.4 ml. A sign is wrong in the above explanation. it should be 1.0xV + 20 x 1.0M...

 

[dat_student]

its 20.4 ml. A sign is wrong in the above explanation. it should be 1.0xV + 20 x 1.0M...

i double checked my math. It's right.

"1.0xV + 20 x 1.0M" makes no sense. You need to subtract the # of moles of the base from the # of moles of the acid & then divide by the total volume

 

[allstardentist]

didnt u have 20.2 before or i saw it wrong? we set up our equations differently, i set the V as the additional volume needed and also the numbers are different. my fault.

 

[dat_student]

didnt u have 20.2 before or i saw it wrong? we set up our equations differently, i set the V as the additional volume needed and also the numbers are different. my fault.

no, 20.4. You can only get 20.4 with the following equation:

V - 20
------- = 10 ^ -2
V + 20

 

[allstardentist]

haha u wanna bet? ... try this...
(1V + 10^-7)/(V+40) = 10^-2

and see what u get... where v is the additional volume needed after the it reached pH 7. i know ur way might be simpler but i just did this by intuition.

 

[dat_student]

haha u wanna bet? ... try this...
(1V + 10^-7)/(V+40) = 10^-2

and see what u get... where v is the additional volume needed after the it reached pH 7. i know ur way might be simpler but i just did this by intuition.

Why do you add 10^-7 to V ????!!!!

 

[allstardentist]

because i did this problem in 2 two steps. first step was to add 20 ml to get to ph=7 and then from there, i calculated how much hcl you need to get from ph=7 to ph=2. thats how i did it

 

[allstardentist]

dat student, do u go to ucla undergrad or ucla dental school alrdy, or you want to go to ucla dental school?

 

[dat_student]

because i did this problem in 2 two steps. first step was to add 20 ml to get to ph=7 and then from there, i calculated how much hcl you need to get from ph=7 to ph=2. thats how i did it

"the 10^-7 part" makes no sense to me. If you want to do it in 2 steps you should use the following equation instead:

(1V)/(V+40) = 10^-2

(i.e. (1V )/( (V+20) + 20) = 10^-2)

That gives you V = 0.4

and then you simply have to add 20 to 0.4 to get 20.4

dat student, do u go to ucla undergrad or ucla dental school alrdy, or you want to go to ucla dental school?

UCLA School of Dentistry Class of 2010 I got in Dec of last yr

 

[allstardentist]

oh i forgot to multiply by 40ml(.040L) to 10^-7. but anyways, you have to add that because at that point the solution is neutral(pH=7). So there is 10^-7(.040) moles of H+ in the solution.

Thats great, im an UCLA undergrad and UCLA is my dream school. hopefully i can get in but we'll see.

 

[allstardentist]

but yeah still get the same answer either way cuz 10^-7 is really small, doesnt affect the answer much.

 

[dat_student]

but yeah still get the same answer either way cuz 10^-7 is really small, doesnt affect the answer much.

true but you can't add it just because it won't affect it much.

(1V)/(V+40) = 10^-2

or

(1V)/((V+20) + 20) = 10^-2

means

(moles of H+ left) divided by the total volume = concentration of 10^-2 M

V+20 is the volume of H+
20 is the volume of OH-

UCLA is a great school. I went to a party school for my other degrees. I partied too much. I'll probably have a hard time adapting to the UCLA lifestyle

 

[allstardentist]

im not saying that... im saying 10^-7 and 10^-7x40ml arent much different since its so small. you have to add 10^-7x40ml term to be truly correct because according to ur equation there is no H+ moles in the solution at ph=7 which is not true. you have to consider the number H+ moles there are at ph=7.

Although i hear ucla dental school is tough, you should be able to handle it..

 

[dat_student]

im not saying that... im saying 10^-7 and 10^-7x40ml arent much different since its so small. you have to add 10^-7x40ml term to be truly correct because according to ur equation there is no H+ moles in the solution at ph=7 which is not true. you have to consider the number H+ moles there are at ph=7.

my 2nd equation is for pH = 2 (i.e. conc. = 10^-2) NOT pH = 7. At pH = 7, we have no extra H+

Although i hear ucla dental school is tough, you should be able to handle it..

I'll do my best

 

[allstardentist]

why is solution pH =7? its because at ph =7, the concentration of H+ in the solution is 10^-7M. This means that 40 ml of solution will have (10^-7M*40ml) moles of H+. Maybe u dont understand what i did exactly,

#of moles of H+ at ph=7 in the solution-> (10^-7Mx40ml) + (1MxV)<-#of H+ that will be added ------------------------
V + 40 <- u know that

that term equals 10^-2M

 

[dat_student]

why is solution pH =7? its because at ph =7, the concentration of H+ in the solution is 10^-7M. This means that 40 ml of solution will have (10^-7M*40ml) moles of H+. Maybe u dont understand what i did exactly,

#of moles of H+ at ph=7 in the solution-> (10^-7Mx40ml) + (1MxV)<-#of H+ that will be added ------------------------
V + 40 <- u know that

that term equals 1.2M

Do we also have this many moles of OH- in the neutral solution: 10^-7 * 40 ml? (i.e. pOH = 7 too) If we do doesn't that cancel 10^-7 * 40 ml H+. I think we need to find the extra H+ right?

 

[allstardentist]

You're right that there is 10^-7x40 ml of OH- in the solution. but you cant say they cancel out because equal number are present. ph=7 means that there are (10^-7x40ml) are dissociated (and same as for OH- as a result). So we need to include it. That's what i think. Does the topscore have your explanation? cuz ihavent taken topscore yet.

 

[dat_student]

You're right that there is 10^-7x40 ml of OH- in the solution. but you cant say they cancel out because equal number are present. ph=7 means that there are (10^-7x40ml) are dissociated (and same as for OH- as a result). So we need to include it. That's what i think. Does the topscore have your explanation? cuz ihavent taken topscore yet.

I don't have topscore. Now, I think you are absolutely right. We usually ignore 10^-7 because it's usually a small number (relatively). In this case (and most other cases), 10^-7 doesn't affect the final answer much but technically & scientifically you're right. Thanks for the explanation.

Here is an interesting problem:

We add 10^-8 moles of H+ to 1 L solution. What is the pH? (assuming volume stays at 1 L)

 

[allstardentist]

assuming the 1L solution is at ph=7, then the pH of the new solution would be -log (10^-7+10^-8)/1. if theres some trick to it, this would probably be wrong.

 

[dat_student]

assuming the 1L solution is at ph=7, then the pH of the new solution would be -log (10^-7+10^-8)/1. if theres some trick to it, this would probably be wrong.

 

[DonExodus]

C1V1=C2V2 C=[] V=Volume.

AT pH=7: C1=10e-7 V=40mL
AT pH=2: C2=10e-2 V=X (20+x actually, just add 20 to x will suffice).

Therefore, (10e-7)(40)=(.01)(x)

4e-6 = .01x

(4x10^-6)/(.01) = 4e4 = .0004L = .4mL = x

20mL+.4mL = total volume of HCl added = 20.4mL

Thats wonderful, but I goofed: Started with mL, ended with L... somehow Im off by 1k, any clue where?

 

[slayerdeus]

We can also deduce that since it only takes .4 mL of HCl to drop the pH to 2 after reaching the isoelectric point at pH 7, the principle of a strong base titrated with strong acid. You know should know that it doesn't take much to lower the pH drastically after the isoelectric point is reached.