Topscore Gen Chem

[mmoosavi41]

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hey everyone. i was struggling with this problem when i did my topscore practice test the other day. i looked at the explanation and still couldn't understand it. if someone can give a good explanation for it, it would be greatly appreciated. heres the question:

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

A) 10.2 mL

B) 20.2 mL

C) 30.4 mL

D) 35.5 mL

E) None of these

thanks in advance.

 

[reely989]

hey everyone. i was struggling with this problem when i did my topscore practice test the other day. i looked at the explanation and still couldn't understand it. if someone can give a good explanation for it, it would be greatly appreciated. heres the question:

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

A) 10.2 mL

B) 20.2 mL

C) 30.4 mL

D) 35.5 mL

E) None of these

thanks in advance.

This was a weird one. Ok, I'll try my best to explain it. Immediately, you recognize that you need 20 mL of HCL to neutralize the 20 mL of NaOH since they are the same concentration. From there, you need to figure out how much 1M HCL more to add to lower the pH to 2. Since the pH is 2, you know that the concentration of H is going to be 10^-2 or .01M. From there, we can setup an equation that looks like:

1M(xmL)=.01M(40+xmL) since we don't know how much to add, only that our final volume will be 40mL plus whatever was added. This is the most difficult part to reason out. But the x simply represents the amount of 1M HCl to be added to the 40mL of total volume. That gives x=.4mL. From there, you know the total volume is 40.4mL, but we don't care about NaOH, so we subtract the 20mL of NaOH and get a final answer of 20.4mL. I hate this question because of the fact that 20.2mL is so close to the correct answer.

I hope this helps. It's tough to follow I know. Maybe someone with a better chemistry mind can come in and save the day.

 

[mmoosavi41]

This was a weird one. Ok, I'll try my best to explain it. Immediately, you recognize that you need 20 mL of HCL to neutralize the 20 mL of NaOH since they are the same concentration. From there, you need to figure out how much 1M HCL more to add to lower the pH to 2. Since the pH is 2, you know that the concentration of H is going to be 10^-2 or .01M. From there, we can setup an equation that looks like:

1M(xmL)=.01M(40+xmL) since we don't know how much to add, only that our final volume will be 40mL plus whatever was added. This is the most difficult part to reason out. But the x simply represents the amount of 1M HCl to be added to the 40mL of total volume. That gives x=.4mL. From there, you know the total volume is 40.4mL, but we don't care about NaOH, so we subtract the 20mL of NaOH and get a final answer of 20.4mL. I hate this question because of the fact that 20.2mL is so close to the correct answer.

I hope this helps. It's tough to follow I know. Maybe someone with a better chemistry mind can come in and save the day.

thanks a lot for the explanation. i'm a little confused with the part where you set up the equation of 1M(xmL)=.01M(40+xmL) though. i understand every thing else though. thanks again

 

[reely989]

thanks a lot for the explanation. i'm a little confused with the part where you set up the equation of 1M(xmL)=.01M(40+xmL) though. i understand every thing else though. thanks again

Well, to me, you basically look at it as C1V1=C2V2. Only, in this case you don't really know either volume. You just know the concentrations, and that the final volume will be whatever you add + 40mL. So, whatever you add (x) of 1M can be related to total volume since you know that you are adding the amount to 40mL already present. I hope this clears it up.

 

[mmoosavi41]

Well, to me, you basically look at it as C1V1=C2V2. Only, in this case you don't really know either volume. You just know the concentrations, and that the final volume will be whatever you add + 40mL. So, whatever you add (x) of 1M can be related to total volume since you know that you are adding the amount to 40mL already present. I hope this clears it up.

okay, great. i think i understand it much better now. thanks again!