titration problem

[Dencology]

How do you approach 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?

so if to neutralize this solution we need 20 ml of HCl. now to get a pH of 2 we need more that 20 ml. how do we do this?
(please show your work)

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[doc toothache]

How do you approach 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?

so if to neutralize this solution we need 20 ml of HCl. now to get a pH of 2 we need more that 20 ml. how do we do this?
(please show your work)

http://forums.studentdoctor.net/showthread.php?p=6800116#post6800116

but change molarity

 

[Glycogen]

So,is the answer 24.2?
4.2 volume of HCl added to initial V of NaOH 20?

 

[doc toothache]

How do you approach 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?

so if to neutralize this solution we need 20 ml of HCl. now to get a pH of 2 we need more that 20 ml. how do we do this?
(please show your work)

Total volume at neutralization=40ml. Now the question becomes what volume of 0.05M HCl is needed to have a solution of 0.01M.

M1V1=M2V2
(0.01)(40)=0.05V2
V2=8ml
Total volume= 20+8=28ml

 

[Mstoothlady2012]

Total volume at neutralization=40ml. Now the question becomes what volume of 0.05M HCl is needed to have a solution of 0.01M.

M1V1=M2V2
(0.01)(40)=0.05V2
V2=8ml
Total volume= 20+8=28ml

how did you get this? I never understand these problems.

 

[Danny289]

how did you get this? I never understand these problems.

think this way first acid must react with whole base that we are going to have
M1V1=M2V2
.05 X V1= .05 X20
and you know it will be 20 ( I think you could guess that must be same as KOH ) now total is 40 ml
pH=2===> N=10^-2=====> M= .01
means how much we have add .05 HCl to 40 ml soultion for having .01 Mole acid:
then
.05 X V=.01 X40
V=8ml
20+8=28 ml total
hope it helped

 

[cougarblue84]

I just ran into that same question on a practice test I just took. After missing the problem I went back to look at it, and I think the solution was more confusing than the question itself, so I'm glad you guys can explain it better. This makes a bit more sense. Thanks.

 

[Mstoothlady2012]

think this way first acid must react with whole base that we are going to have
M1V1=M2V2
.05 X V1= .05 X20
and you know it will be 20 ( I think you could guess that must be same as KOH ) now total is 40 ml
pH=2===> N=10^-2=====> M= .01
means how much we have add .05 HCl to 40 ml soultion for having .01 Mole acid:
then
.05 X V=.01 X40
V=8ml
20+8=28 ml total
hope it helped

got it! thanks

 

[Danny289]

got it! thanks

if you got the concept solve this problem with H2SO4

 

[TexasDent]

if you got the concept solve this problem with H2SO4

You would have to take into account the 2 hydrogens liberated by one molecule of H2SO4 which is 2 Normality. So you just multiply the molarity given by 2 and use that as the molarity for H2SO4. Everything else is the same.

 

[drgreen]

As you said the first part is to neutralize the solution with 20 mL HCl.

Next we set up an M1V1=M2V2, look at it as a dilution:
we are taking X mL of .05 HCl solution, and we are diluting it with the 40 mL of neutralized solution (where [HCl]=[KOH]) to get a final volume of 40+X mL.

Therefore...
M1=.05 V1=X
M2=.01 (desired) V2=40+x

Plug into equation...
(.05)(X)=(.01)(40+X)
.05X=.4+.01X
.04X=.4
X=10 mL

Since we added 20 mL to Neutralize the KOH, then we added 10 mL to create pH of 2
We added a total of 30 mL of HCl.

WORK BACKWARDS TO CHECK...
We added a total of 30 mL HCl and 20 mL KOH for a total of 50 mL solution.
20 mL .05 M HCl neutralize 20 mL of .05M KOH.
Now we have 40 mL of neutral solution, and 10 mL of .05M HCl remaining.
The concentration (mol H+/L) is what were looking for to get pH.
we have 5*10^-4 mol HCl (0.05mol/L*0.01L) in 50 mL solution.
overall [H+]=(5*10^-4 mol)/(.050 L solution)=1*10^-2
pH=-log(1*10^-2)=2

i hope this helps

 

[Mstoothlady2012]

if you got the concept solve this problem with H2SO4

Okay! let me try. I know we have to do normality here so..

NaVa = NbVb
0.1 x Va = 0.05 x 20ml
Va = 10 ml <-- this much of 0.05 H2SO4 needed to neutralize 20 ml of 0.05 KOH

So now we have total volume = 30ml
Solve for = how much of 0.05 H2SO4 needed to make 30ml of solution of pH=2

M1V1 = M2V2
(0.01)(30) = (0.05)V2
V2 = 6

We need a total of 16 ml of H2SO4...is this right?

 

[drgreen]

Okay! let me try. I know we have to do normality here so..

NaVa = NbVb
0.1 x Va = 0.05 x 20ml
Va = 10 ml <-- this much of 0.05 H2SO4 needed to neutralize 20 ml of 0.05 KOH

So now we have total volume = 30ml
Solve for = how much of 0.05 H2SO4 needed to make 30ml of solution of pH=2

M1V1 = M2V2
(0.01)(30) = (0.05)V2
V2 = 6

We need a total of 16 ml of H2SO4...is this right?

lets check by working backwards...

now we have 20 mL of .05 M KOH and 16 mL of .05 M H2SO4 for a total of 36 mL, right?

Total mol of H+:

(0.05mol/L HCl)(.016 L HCl)(2 mol H+/1 mol HCl)=0.0016 mol H+

Total mol of OH-:

(0.05 mol/L)(.02 L)=0.001 mol OH-

now, .001 mol of OH- are neutralized with .001 mol of H+
this leaves us with (.0016-.001) .0006 mol of H+.
remember we have 36 mL of solution.

So our concentration of H+ is calculated as follows...
.0006mol/.036L=1.66*10^-2
our pH=a bit less than 2.

please look at my previous answer to see where this discrepancy arises, i'm really sorry if this is confusing at all.

 

[Danny289]

Okay! let me try. I know we have to do normality here so..

NaVa = NbVb
0.1 x Va = 0.05 x 20ml
Va = 10 ml <-- this much of 0.05 H2SO4 needed to neutralize 20 ml of 0.05 KOH

So now we have total volume = 30ml
Solve for = how much of 0.05 H2SO4 needed to make 30ml of solution of pH=2

M1V1 = M2V2
(0.01)(30) = (0.05)V2
V2 = 6
We need a total of 16 ml of H2SO4...is this right?

Bravo, first part excellent. but the second part you forgot something you did in first part you have to go by
NV=N1V1 equation not MV!( exact same you did in first part you are working with H2SO4 not HCl) then:
.05 X 2=0.1
.01 X30= .1 XV=========> V=3
10 +3= 13 ml H2SO4

 

[Mstoothlady2012]

Bravo, first part excellent. but the second part you forgot something you did in first part you have to go by
NV=N1V1 equation not MV!( exact same you did in first part you are working with H2SO4 not HCl) then:
.05 X 2=0.1
.01 X30= .1 XV=========> V=3
10 +3= 13 ml H2SO4

uh! I thought we don't do Normality again since we did it already once. Okay thanks!

 

[doc toothache]

Next we set up an M1V1=M2V2, look at it as a dilution:
we are taking X mL of .05 HCl solution, and we are diluting it with the 40 mL of neutralized solution (where [HCl]=[KOH]) to get a final volume of 40+X mL.

Therefore...
M1=.05 V1=X
M2=.01 (desired) V2=40+x

Plug into equation...
(.05)(X)=(.01)(40+X)
.05X=.4+.01X
.04X=.4
X=10 mL

Since we added 20 mL to Neutralize the KOH, then we added 10 mL to create pH of 2
We added a total of 30 mL of HCl.

i hope this helps

When the solution added is of low concentration it's volume needs to be taken into account.

 

[Mstoothlady2012]

Thanks Dr. green for pointing out my mistake. You did a very good job at explaining. Let me try this H2SO4 problem one more time.

How many milliliters of 0.05 M H2SO4 are required to turn 20 ml of 0.05 M KOH into a solution of pH 2.00?

NaVa = NbVb

0.01 x Va = 0.05 x 20ml
Va = 10 ml <--- this much needed to neutralize 20 ml of 0.05 KOH

20 + 10 = 30 ml <--- volume after neutralizing (not the total volume)

N1V1 = N2V2
(0.01)(30+x) = (0.1)X
0.3 + 0.01x = 0.1 x
0.3 = 0.09x
x = 3.33

Total of 13.33 ml of H2SO4 is needed to turn 20 ml of 0.05 M KOH into a solution of pH 2.00

Now let's double check

0.05 mol/L x 0.02 L = 0.001 mol of OH
0.05 mol/L x 0.0133L x 2 = 0.00133 mol of H+

Since we 0.001 mol of H+ was used to neutralize 0.001 mol of OH we are left with...

mol of H+ left = 0.00133 - 0.001 = 3.3 x 10^ -4 mol

[H+] = 3.3 x 10^ -4 mol/ 0.033 L
= 0.01
= 1 x 10^ -2

pH = - (2 - log 1) = 2

I hope I did it right this time!

 

[Danny289]

Thanks Dr. green for pointing out my mistake. You did a very good job at explaining. Let me try this H2SO4 problem one more time.

How many milliliters of 0.05 M H2SO4 are required to turn 20 ml of 0.05 M KOH into a solution of pH 2.00?

NaVa = NbVb

0.01 x Va = 0.05 x 20ml
Va = 10 ml <--- this much needed to neutralize 20 ml of 0.05 KOH

20 + 10 = 30 ml <--- volume after neutralizing (not the total volume)

N1V1 = N2V2
(0.01)(30+x) = (0.1)X
0.3 + 0.01x = 0.1 x
0.3 = 0.09x
x = 3.33

Total of 13.33 ml of H2SO4 is needed to turn 20 ml of 0.05 M KOH into a solution of pH 2.00

Now let's double check

0.05 mol/L x 0.02 L = 0.001 mol of OH
0.05 mol/L x 0.0133L x 2 = 0.00133 mol of H+

Since we 0.001 mol of H+ was used to neutralize 0.001 mol of OH we are left with...

mol of H+ left = 0.00133 - 0.001 = 3.3 x 10^ -4 mol

[H+] = 3.3 x 10^ -4 mol/ 0.033 L
= 0.01
= 1 x 10^ -2

pH = - (2 - log 1) = 2

I hope I did it right this time!

yes you did right this time