chem question I HATE THESE PROBLEMS

[pistolpete007]

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

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[osimsDDS]

(20ml NaOH)(.01 pH)= (1M HCl)(v)
v= .2
true volume = 20 + 0.2 = 20.2 added to NaOH solution to obtain pH of 2

Or you can do it another way....

find out the volume to neutralize the reaction (you know how to do that):

you will get 40ml...

(1M HCl)(x)= (.01 pH)(40 + x)
1x=.4 + .01x
.99x= .4
x= around .2
40 - 20= 20 ml +.2 = 20.2ml of HCl

hope that helps

 

[pistolpete007]

(20ml NaOH)(.01 pH)= (1M HCl)(v) why is the .01pH assoc. with NaOH when the pH is supp to be of the final solution?
v= .2
true volume = 20 + 0.2 = 20.2 added to NaOH solution to obtain pH of 2

Or you can do it another way....

find out the volume to neutralize the reaction (you know how to do that):

you will get 40ml...wouldnt u get 20 bc 1N HCL x =1N NaOH*20mL so x=40?

(1M HCl)(x)= (.01 pH)(40 + x)
1x=.4 + .01x
.99x= .4
x= around .2
40 - 20= 20 ml +.2 = 20.2ml of HCl

hope that helps
__________________

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2009 cycle sGPA: 3.82 cumGPA: 3.72 DAT: N/A yet

 

[osimsDDS]

yea you would get 20 ml but you have to add it to 20 ml of existing NaOH, so its 40 (20+20=40)

As for the first part, thats how ppl have been doing it lately, but i use my own method, the 2nd method i use because as it threw you off it threw me off as well when i did it also, most ppl do it the second way...its easier to understand but the calculations may be hard towards the end if you have weird numbers...

 

[doc toothache]

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

The total volume of the solution at neutralization is 40ml. Then the question becomes what volume is needed to have (40ml) of pH 2 solution.

M1V1= M2V2
(0.01)(40)=(1)V2
V2=0.4ml

Total volume of HCl= 20 + 0.4= 20.4ml

 

[theedaddy77]

osim u had the right strategy but you calculated wrong. the way you do it is the way i like to do it also. when you said .99x=.4 somehow u conclused that x =.2 but should be .4 therefore u would end up adding 20.4 ml of HCL