titration question

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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?

This is a question off of TopScorePro.
They provide you with an explanation for the answer, but I don't understand where they are got .01 M in the first equation.?

TopScore's explanation wrote:
(1M)(x mL) = (.01M){(40+x mL)}
x = 0.4 + 0.1x
0.99x = 0.4
x = 0.4 mL
therefore,
40.4 - 20 mL = 20.4
Answer: 20.4 mL
 
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.01 is 10^-2. This is the H+ concentration.
 
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ooh right right because -log(0.01) is 2
gotcha!

Okay so one more question, I'm confused on what is what in the first equation.
The first half of the equation says (1M)(x mL), is that referring to 1M of NaOH or 1M of HCL or..?
I guess I need a explanation of how this equation works to find the volume of HCL needed, because with just the numbers I can't seem to follow.
Thanks!
 
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To reach eq there are 20 ml of HCl and 20 ml of NaOH which give 40 ml overall. The equation is essentially M1V1=M2V2. So you are finding out how much more than twenty ml of HCl you must add to reduce the solution from neutral to a pH of 2.

Thus you have a 1M solution of HCl and you want to know how much is needed to essentially change the concentraiton to 0.01M (the H+ concentration) When you solve for X this is the amount above the original 20ml that is needed.

I hope this is clear I am not good at explaining myself.
 
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PD3 said:
To reach eq there are 20 ml of HCl and 20 ml of NaOH which give 40 ml overall. The equation is essentially M1V1=M2V2. So you are finding out how much more than twenty ml of HCl you must add to reduce the solution from neutral to a pH of 2.

Thus you have a 1M solution of HCl and you want to know how much is needed to essentially change the concentraiton to 0.01M (the H+ concentration) When you solve for X this is the amount above the original 20ml that is needed.

I hope this is clear I am not good at explaining myself.
Click to expand...


Hi, I understand where you are getting the 40 mL by using the formula MV=M2V2 but I don't see how you know how to set up the second formula to find the amount of HCL needed to be added. Is this a specific set up formula used for titration problems? if so is this how you do all titration problems acid and basic?Any clarification would be much appreciated
 
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The first important thing to realize here is that there is a strong acid and a strong base with the same molarity. Therefore, when equal amounts (moles) are added the solutions is at equilibrium. Here it would be seven since it is a strong acid and a strong base. Think of it as this, we need to find the excess moles of H+ which is equal to the moles of H+ minus the moles of OH-.

Since we know how much to add in order to find the equilibrium, we only have to solve for the additional amount needed to reach a pH of 2 (the second equation you speak of). In a sense we are solving it as if there was no base present because we are going from 7 to 2. Since a pH of 2 gives a [H+] of .01 we know what concentration we need to solve for, thus the equation.

This is not a specific formula and it is not typically how I would solve them. In chem class you would typically set up an ICE table or something, but it isn't necessary here.
 
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oh okay, so we're using the formula m1v1=m2v2

but why wouldn't v1 be 40 ml? Since you are starting off at 40 ml from the 20 ml of HCL and 20 ml of NaOH.
 
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Because that is the volume we are going to add to the solution after the equivalence point. M1V1=M2V2 is saying that at they are equal, thus we need to know what volume added of HCl makes them equal. We know in the end we will have 40mL total plus some amount. THat amount has to be the amount of HCl we are adding.

Does that make more sense? At this point, I think it would be safe to give the example that we already have 40 mL of water and we want to know what volume of 1M HCl will give us a pH of 2. The equation would be set up the same way. ONly at the end we would not add the amount to 20 mL for the answer.
 
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ooh I completely get it now.
Thank you so much for your help, I really appreciate it!
 
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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?
E. 30
The answer is E.



This is what I found from search for titration problem.
I just thought that adding this problem here could help you studying 😀


HCl and KOH have the same molarity so adding 20ml of HCL will make the solution into neutral.
Now the fun begins.
M1V1 = M2V2
(0.05M)(x) = (0.01M)(40ml + x)
0.05x = 0.4 + 0.01x
0.04x=0.4
x= 10 mL
therefore,
50ml - 20ml = 30
 
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