gchem question

dontbam · Jul 18, 2006

Can anyone help me with this question?

What volume of HCl was added if 20mL of 1M NaOh is titrated with 1M of HCl to produce a pH=2 solution?

according to topscore the answer is 20.4 mL

tissy · Jul 18, 2006

Think this is right..if not sorry...😛

thehipster · Jul 18, 2006

The answer doesn't happen to be 200 mL, does it?

asckwan · Jul 18, 2006

I got it to be .2L. Anyone else?

(forgot my dot there 😳 )

lovechild404 · Jul 18, 2006

i got the answer to be .2L = 20 mL. i'm not sure why the answer would be 20.4mL...where did the .4 come from?

dontbam · Jul 19, 2006

hey can you guys show me how you calculated the 20ml? in top score they set it up where it's like 0.01 * (40 + x) = x

solving, that's where the .4 comes from, how did you guys solve it?

Chga · Jul 19, 2006

you have to convert to normality
1 mole NaOH/1L = x mole NaOH/.02L
x= .02 moles
(.02 moles)(1 equiv of -OH)= .02 N
since OH and H in HCl are 1 to 1 to equalout
there is .02N of H+ ions meaning .02M of HCl
since its 1M HCl there are .02L of it.
.02L = 20ml
have no idea how some people got .2L or where the 20.4 comes in.
it doesnt make sense

djeffreyt · Jul 19, 2006

dontbam said:
Can anyone help me with this question?

What volume of HCl was added if 20mL of 1M NaOh is titrated with 1M of HCl to produce a pH=2 solution?

according to topscore the answer is 20.4 mL

first...you have 20mL of a 1M Strong Base that will take only 1 proton per molecule (NaOH). To neutralize that, add an equal amount of a strong acid (HCl) at 1 M. So to neutralize the 20mL of 1 M NaOH, we add 20 mL of 1 M HCl.

That was step 1, and you do that in your head just by looking at the problem. Then the question is how much more do we have to add of 1M HCl to our now 40mL of neutral solution to get the pH = 2

You need to know this formula below and how to work with it
pH = 2
pH = -log [H+]
2 = -log [H+]
10^-2 = [H+]
or
[H+] = 0.01 M

if we call X the amount of HCl we need to add to our 40ml solution to reach a pH = 2 then you can put it together like this. It's kind of a dilution question. You start with 1 M HCl. You are diluting it by adding some of this 1M HCl to a 40mL neutral solution. When will the solution's pH drop of 0.01 M HCL 9which is pH =2)

(1 M HCl) * (X mL) = (0.01 M HCl) * (40 + X ml)

I'm not gonna do the math here...but the answer comes out right.

asckwan · Jul 19, 2006

Oh dang! I see now, 20mL is the amount to neutralize the solution, but then you have to take it one step further to get the pH = 2.

Achiever also has a question like this:

How many milliliters of 0.05 M HCl are required to turn 20 ml of 0.05 M KOH into a solution of pH 2.00?

Practice practice practice!!

dontbam · Jul 19, 2006

thanks guys!

fancymylotus · Jul 19, 2006

asckwan said:
Oh dang! I see now, 20mL is the amount to neutralize the solution, but then you have to take it one step further to get the pH = 2.

Achiever also has a question like this:

How many milliliters of 0.05 M HCl are required to turn 20 ml of 0.05 M KOH into a solution of pH 2.00?

Practice practice practice!!

Can someone do this problem out? I'm having a ****** moment.

cocokeala · Jul 19, 2006

finishing up Dr. Jeffs math hope its right cuz i came out with the right answer 🙂

V= .4+.01V
.99V= .4
V= .4

.4 is the extra amount of HCL you need to add to the neutralized soln which was 40 (remember 20 HCL + 20NaOH)

well we only want how much HCL was added
so we know that 20 HCL was added just to neutralize the NaOH, and we added .4 HCL to get it to pH=2.

so 20+.4= 20.4

lovechild404 · Jul 20, 2006

BRILLIANT!...we're all on our way to dental school now! :laugh:

tom_servo_dds · Jul 20, 2006

amsie said:
Can someone do this problem out? I'm having a ****** moment.

I came up with 30 mL for HCl needed:

(0.05M HCl)(X mL) = (0.01)(40mL + X mL)
X = 10 mL + 20mL to neutralize = 30 mL total

fancymylotus · Jul 20, 2006

tom_servo_dds said:
I came up with 30 mL for HCl needed:

(0.05M HCl)(X mL) = (0.01)(40mL + X mL)
X = 10 mL + 20mL to neutralize = 30 mL total

huh. 40mL 😕

tom_servo_dds · Jul 20, 2006

amsie said:
huh. 40mL 😕

djeffreyt had a good explanation above where the 40mL comes from. Also, this explanation may help clear some of it up too: http://forums.studentdoctor.net/showthread.php?t=297575

Envision · Aug 8, 2006

tom_servo_dds said:
I came up with 30 mL for HCl needed:

(0.05M HCl)(X mL) = (0.01)(40mL + X mL)
X = 10 mL + 20mL to neutralize = 30 mL total

Hey tom,

doing that problem i got 28. is your 30 round off? or is my math incorrect? thanx.

dat_student · Aug 8, 2006

dontbam said:
Can anyone help me with this question?

What volume of HCl was added if 20mL of 1M NaOh is titrated with 1M of HCl to produce a pH=2 solution?

according to topscore the answer is 20.4 mL

ok, I solve it again:

V is the total volume of HCl

1 * V - 1 * 20
__________ = 10^-2
V + 20

V = 20.2 / 0.99

V = 22.4

or you can do the following:

20 ml of HCl is needed to neutralize 20 ml of 1 M NaOH. So, total volume of the final solution is V + 20 + 20 (or V + 40). V is the extra volume of HCl that is needed to produce a pH=2 solution.

1 * V
_______ = 10^-2
V + 40

V = 0.4

20 (to neutralize) + 0.4 (extra) = 20.4

*Again, I assumed that at pH = 7 we have little H+.

tom_servo_dds · Aug 8, 2006

Envision said:
Hey tom,

doing that problem i got 28. is your 30 round off? or is my math incorrect? thanx.

You're right, 28 - thanks for catching that!

Envision · Aug 9, 2006

tom_servo_dds said:
You're right, 28 - thanks for catching that!

no wait, your right! i did it again and i got thirty! haha 😀

tom_servo_dds · Aug 9, 2006

Envision said:
no wait, your right! i did it again and i got thirty! haha 😀

Looks like we both made the same mistake :laugh:

- it is 30, which is good since the options were:

A. 10
B. 15
C. 20
D. 25
E. 30

gchem question

Member

Full Member

Senior Member

Member

Member

Member

Member

Senior Member

Member

Member

A Whole New World

katonk!

Member

Senior Member

A Whole New World

Senior Member

Envisioning...

Junior Member

Senior Member

Envisioning...

Senior Member

Similar threads