Titration Question

[BeachBumDMD]

Advertisement - Members don't see this ad

Hey Everyone. I saw this question was asked before in a thread a few years ago but their thread didn't help me understand where a certain amount came from.

If 10 mL of 1M NaOH is tritated with 1 M HCL to a pH of 2, what volume of HCl was added??

I understand that to neutralize the solution, it'll take 10 mL of HCl. However, the next step in finding the final amount to make the pH 2, it uses this equation:

(1M)(x mL) = (0.01 M)[(20 +x)mL]

Where did they get the 0.01M from????

 

[Kneecoal]

i'm not really sure either, but maybe the fact that 10ml = 0.01 L has something to do with it?

 

[nze82]

Hey Everyone. I saw this question was asked before in a thread a few years ago but their thread didn't help me understand where a certain amount came from.

If 10 mL of 1M NaOH is tritated with 1 M HCL to a pH of 2, what volume of HCl was added??

I understand that to neutralize the solution, it'll take 10 mL of HCl. However, the next step in finding the final amount to make the pH 2, it uses this equation:

(1M)(x mL) = (0.01 M)[(20 +x)mL]

Where did they get the 0.01M from????

HCl is a strong acid; therefore, it dissociates 100% via the following formula:
HCl --> H+ + Cl-

We know that:

pH = -log[H+] --> 2 = -log[H+] -->[H+] = 10^-2 = 0.01

Now that you have the concentration of H+ and you know that HCl is a strong acid that dissociates 100% you can work backwards and find that the concentration of [HCl] = 0.01, if the pH of the solution is pH = 2.

Now we can rephrase the above question like this:
How much of a 1M HCl do we need to add to our solution so that the final [HCl] = 0.01

Now use the following formula to find the answer:

M1V1 = M2V2

(1M)(Xml) = (0.01M)(20+Xml)

NOTE: It's 20 + X ml, because we have already added 20ml to neutralize the solution, and now we're adding an additional X ml to satisfy the condition that the question is asking for.

Now simply solve for X.

Hope this helped!

 

[Kneecoal]

nze, just to clarify things:

when using M1V1 = M2V2 in a regular titration question, one side doesn't have the total added volume - each side represents that acid/base's volume.

but when we're satisfying a condition past neutralization, then one side represents the total added volume?

 

[nze82]

nze, just to clarify things:

when using M1V1 = M2V2 in a regular titration question, one side doesn't have the total added volume - each side represents that acid/base's volume.

but when we're satisfying a condition past neutralization, then one side represents the total added volume?

In either case, you're always determining the total volume of the acid/base (not the total volume of the solution) that's been added to satisfy a particular condition. In other words, you're NEVER solving for the total volume of the solution; You're ALWAYS solving for the total volume of the acid/base added.
Think of it as a two-step process:

First step: You want to find the total volume of the acid that needs to be added to neutralize the solution. We found this volume to be 20ml.

Second step: You want to find the total volume of the acid that needs to be added to have a solution of pH = 2.

 

[Kneecoal]

In either case, you're always determining the total volume of the acid/base (not the total volume of the solution) that's been added to satisfy a particular condition. In other words, you're NEVER solving for the total volume of the solution; You're ALWAYS solving for the total volume of the acid/base added.
Think of it as a two-step process:

First step: You want to find the total volume of the acid that needs to be added to neutralize the solution. We found this volume to be 20ml.

Second step: You want to find the total volume of the acid that needs to be added to have a solution of pH = 2.

ok, that's where i'm getting confused. i thought it was originally 10 ml of 1M HCl to neutralize 10 ml of NaOH like beachbum said. sorry if i'm being thick, but where exactly are you getting 20 from? thanks for all the help.

 

[PooyaH]

ok, that's where i'm getting confused. i thought it was originally 10 ml of 1M HCl to neutralize 10 ml of NaOH like beachbum said. sorry if i'm being thick, but where exactly are you getting 20 from? thanks for all the help.

20ml is your new final volume. Remember that you already had 10ml of base and now you have added 10ml acid to neutralize it, so you have a total of 20ml so far! But you still need to add some more acid for the pH to get to 2, so it's represented as 20+x