solve this question please

[deeentist79]

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this Gen chem question from top score is puzzling, can someone explain this?

Q: what volume of HCL was added if 20 ml of 1 M NaOH is titrated with 1 M HCl to produce a PH = 2 ???

 

[412smb412]

this Gen chem question from top score is puzzling, can someone explain this?

Q: what volume of HCL was added if 20 ml of 1 M NaOH is titrated with 1 M HCl to produce a PH = 2 ???

20 ml of 1 M NaOH with 20 ml of 1 M HCl is a neutralization therefore pH for this will be 7. so to bring the pH down to 6 it would be to multiply by 10 -> 20 ml x 10.... 7-2 = 5. therefore it should contain 10^5 more.

20 mL x (10^5)

 

[aqz]

SMB is right on the neutralization part...but I think he forgot a few things.

This question is often asked, and there are a few traps you need to watch out for.

Since they're both 1 M, you'd need 20 ml of HCl to reach a pH of 7, e.g. neutral.

Here's the 1st trap: after you add 20 ml of HCl, the total volume is 40 ml. So when you do M1V1 = M2V2, the volume is no longer 20 ml.

You need to produce a pH of 2...which is a H+ concentration of....

2 = -log[H+]

[H+] = 10^(-2) = 0.01 M

So if you're adding another "x" volume IN ADDITION to the 20 ml you already added....using M1V1 = M2V2

x*1 M = (40+x) * 0.01 M

x = 0.4/0.99 = 0.409

So then add that onto the original amount that you added (20 ml) and you get 20.409 ml.

 

[412smb412]

SMB is right on the neutralization part...but I think he forgot a few things.

This question is often asked, and there are a few traps you need to watch out for.

Since they're both 1 M, you'd need 20 ml of HCl to reach a pH of 7, e.g. neutral.

Here's the 1st trap: after you add 20 ml of HCl, the total volume is 40 ml. So when you do M1V1 = M2V2, the volume is no longer 20 ml.

You need to produce a pH of 2...which is a H+ concentration of....

2 = -log[H+]

[H+] = 10^(-2) = 0.01 M

So if you're adding another "x" volume IN ADDITION to the 20 ml you already added....using M1V1 = M2V2

x*1 M = (40+x) * 0.01 M

x = 0.04/0.99 = 0.0409

So then add that onto the original amount that you added (20 ml) and you get 20.0409 ml.

dam. ur right! but how did u go from: x*1 M = (40+x) * 0.01 M to x = 0.04/0.99 = 0.0409?


isn't it 0.4/0.99 + 20

 

[aqz]

Guess I made a mistake too, nice catch.

Edited my post.

 

[412smb412]

Guess I made a mistake too, nice catch.

Edited my post.

can't believe i made that mistake in the first place 😡

 

[toyotasupra]

Just remember, the REAL DAT has a selection of answers. You can always plug and play to see which one works IF you are in a real bind. Hehe, I remember doing that my first time. Hopefully I don't have to resort to such things tomorrow as it will burn up precious minutes. 😀

 

[deeentist79]

THanks a lot folks, yah the correct answer is 20.4 ml.

 

[stpele01]

SMB is right on the neutralization part...but I think he forgot a few things.

This question is often asked, and there are a few traps you need to watch out for.

Since they're both 1 M, you'd need 20 ml of HCl to reach a pH of 7, e.g. neutral.

Here's the 1st trap: after you add 20 ml of HCl, the total volume is 40 ml. So when you do M1V1 = M2V2, the volume is no longer 20 ml.

You need to produce a pH of 2...which is a H+ concentration of....

2 = -log[H+]

[H+] = 10^(-2) = 0.01 M

So if you're adding another "x" volume IN ADDITION to the 20 ml you already added....using M1V1 = M2V2

x*1 M = (40+x) * 0.01 M

x = 0.4/0.99 = 0.409

So then add that onto the original amount that you added (20 ml) and you get 20.409 ml.

Instead of 1 M of HCl, what if we had 2 M of HCl? How would it affect the result?

 

[aqz]

Instead of 1 M of HCl, what if we had 2 M of HCl? How would it affect the result?

Uh.

Volume needed to neutralize in the beginning would be 10 ml HCl instead of 20. Plug and chug that like how I did in my post above, think it should be pretty clear.

 

[stpele01]

Uh.

Volume needed to neutralize in the beginning would be 10 ml HCl instead of 20. Plug and chug that like how I did in my post above, think it should be pretty clear.

Thank you for the help. just checking.