How do you solve this pH problem?

[Pentako]

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Can someone explain this problem in laymen terms??

Thanks in advance,

Topscore #2 problem 68

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

 

[license2kill]

Reason for editing: to save face, my explanation made no sense😳
Sorry guys!

 

[atlanta213]

anyone here?

 

[osimsDDS]

this question is infamous...

pH of 2 = 1x10^-2

Neutralization between the HCl and the NaOH will give you 20ml, add that to 20ml you started out with and you get 40 ml...

Now you use this formula (many people do it differently but i find this the best way for me):

(HCl M)(x) = (pH)(total volume + x)

(1M)(x) = (1x10^-2)(40 + x)
1x= .4 + .01x
x= .4ml

Now add that to the starting volume of HCl which was 20 ml so your answer is 20.4ml

 

[atlanta213]

OsimsDDS

I am proud of you, man!

However, the calculation here is wrong
1x= .4 + .01x
x= .4ml

1x-0.1x=.4
0.9x=4
approximately 4.4

 

[Pentako]

Thanks much!, but where did you derive the formula from?

For example, am I supposed to just memorize it?

 

[osimsDDS]

yup memorize it, i first got it from topscore on one of their tests...just memorize the equation and you will know how to do it every time.

 

[Pentako]

thanks!

 

[atlanta213]

what was the answer ? 20.4? isn't that 24.4?

 

[PooyaH]

.4ml does not make sense! It should be 20.4ml!

When you have 1M NaOH and 1M of HCl you should know that to reach the equil. point which would be 7 since we're adding a strong acid to a strong base, you need to add 20ml of HCl since you have 20ml of NaOH! So we know that the answer should def. be more than 20ml! Osims did the calculation right but forgot to add 4.4 ml to 20ml at the end!

 

[PooyaH]

OsimsDDS

I am proud of you, man!

However, the calculation here is wrong
1x= .4 + .01x
x= .4ml

1x-0.1x=.4
0.9x=4
approximately 4.4

hmmm, I'm sure Osims did it right!
1x = .4 + .01x.....................(1x10^-2 = .01 not .1)
1x - .01x = .4
.99x = .4
x = .4!

 

[drgreen]

Osims is dead on

 

[license2kill]

So, do we just memorize the formula and use it when a volume is asked for a neutralization problem at a certain pH?

I can't seem to conceptualize the logic behind it. Osims, can you explain?

 

[klutzy1987]

OsimsDDS

I am proud of you, man!

However, the calculation here is wrong
1x= .4 + .01x
x= .4ml

1x-0.1x=.4
0.9x=4
approximately 4.4

Not sure how you got 4.4 from these numbers but they equal .4. Osims was correct.

 

[klutzy1987]

Another way to do this that makes sense and I understand the reason behind is like this.

(# of moles of HCl - # of moles of NaOH)/(The total Volume in Liters)=will give you the concentration of H+ after its fully titrated.

Setting this equation up gives (1*x) - (1*.02) = 2*10^-2mol/L*(x+.02)

x is the volume of HCl,
(x+.02) is the total volume

Solving for x gives X - .02 = .02 * (X + .02)

X - .02 = .02x +.0004
.98x = .0204
X = .0208
Covert that to milliliters gives us 20.8 milliliters which is actually the correct answer.

 

[marscole]

at neutral [H+] = [OH] = 1.0 x 10^-7
so... @ pH 2 [H+] will be 1.0 x 10^-2 = 0.01

the volume is 40 ml now. how much still need to add so that the final [H+] is 0.01 M?

the equation is M1V1 = M2V2

(1 M HCl) (V1 (assuming 1 M of HCl is in the buret)) = (0.01 M of final [H+]) (40 mL + V1)

So.... V1 is 0.4040404040..... ml dont forget to add 20 ml initially...😉

dont just memorize the equations... learn to ask yourself why those numbers are settled there....🙂

 

[acupofchai]

nice..i get it now =)

 

[Datgl]

Atlanta,,,,you mistakingly divided 4 by .9, its supposed to be .4/.9=.44