How to go about Strong Acid Strong Base titrations?

[jhanago]

What volume of HCl was added if 20 ml of 1m NAOH is titrated with 1M HCl to produce a ph = 2


I understand that 20 ml of HCl neutralized 20 ml of NaOH... but how do we go about thinking of the pH it made?

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[bharat008]

What volume of HCl was added if 20 ml of 1m NAOH is titrated with 1M HCl to produce a ph = 2


I understand that 20 ml of HCl neutralized 20 ml of NaOH... but how do we go about thinking of the pH it made?

pH = 2 so H+ = .01mol
mol NaOH = .02mol
If you add .03 mol of HCl then you will be left with .01 mol which would give you pH of 2..

Volume of HCl = 30ml
mol HCl added is .03mol = (M)(L) = (1M)(.03L) = molarity is already given to us..

Took me like 5/6 min to figure it out.. Can't think right now...sleepy.

 

[YoonS]

Just got curious,

When you are calculating for the pH you need to get the [H+] right?

but as you say, the concentrartion of H+ will be different from 0.01

since the total volume of the solution would be 50mL.

How do you encounter that???

pH = 2 so H+ = .01mol
mol NaOH = .02mol
If you add .03 mol of HCl then you will be left with .01 mol which would give you pH of 2..

Volume of HCl = 30ml
mol HCl added is .03mol = (M)(L) = (1M)(.03L) = molarity is already given to us..

Took me like 5/6 min to figure it out.. Can't think right now...sleepy.

 

[bharat008]

Just got curious,

When you are calculating for the pH you need to get the [H+] right?

but as you say, the concentrartion of H+ will be different from 0.01

since the total volume of the solution would be 50mL.

How do you encounter that???

They already gave us Molarity of HCl and we found the moles that we added so i just solved for Volume..
V = (.03mol)(1M) = .03L

But i know what you mean. Not really sure how you would account for 50mL of total volume.
That was the best guess i could get for that question.

 

[reely989]

Top Score has a different solution for this. They basically say that you know you end up with 40mL after the initial neutralization, and you know you have to add x amount of 1M HCl in order to arrive at a final concentration of .01M and a final volume of 40+x(amount that you added), so we get

1M(x)=(.01M)(40+x). You solve for x, which is about .4mL. You then take the total volume, which is 40.4mL, and substract 20mL from it because you aren't interested in NaOH. I could not follow the reasoning of this problem at all honestly. I was grateful there wasn't a question like this on my DAT.

I believe the more traditional method of solving this would include an ICE table maybe? I'm not sure. I was just ready to make a similar equation to that if I saw a titration like this.

 

[bharat008]

Top Score has a different solution for this. They basically say that you know you end up with 40mL after the initial neutralization, and you know you have to add x amount of 1M HCl in order to arrive at a final concentration of .01M and a final volume of 40+x(amount that you added), so we get

1M(x)=(.01M)(40+x). You solve for x, which is about .4mL. You then take the total volume, which is 40.4mL, and substract 20mL from it because you aren't interested in NaOH. I could not follow the reasoning of this problem at all honestly. I was grateful there wasn't a question like this on my DAT.

I believe the more traditional method of solving this would include an ICE table maybe? I'm not sure. I was just ready to make a similar equation to that if I saw a titration like this.

Never seen a question like this.. It's a good one..