Titration :(

[greenseeking]

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Hi I need help with this titration problem!

how many ml of .010 M NaOH need to be added to take 10 mL 0.010 M lysine from pH=5.68 (the first equivalence point) to pH=10.80 (pKa for lysine)?

a. 5 ml .10 M NaOH
b. 10 ml .10 M NaOH
c. 15 ml .10 M Naoh
d. 20 ml .10 M NaOH

Answer is C

 

[Perpetually]

Liters NaOH added = moles OH titrated / M NaOH

Moles OH titrated = moles H+ neutralized to get to end point of titration, which in this case is provided by the change in pH

Moles of H+ needed to get from first to second pKa = 10^(-5.68) - 10^(-10.8). note how in the difference taken, the higher pH represents the solution with fewer moles of H+ present per liter of solution.

So once you have moles of H+, you have moles of OH needed and then can go through NaOH molarity to find how many ml needed.

 

[Postictal Raiden]

Liters NaOH added = moles OH titrated / M NaOH

Moles OH titrated = moles H+ neutralized to get to end point of titration, which in this case is provided by the change in pH

Moles of H+ needed to get from first to second pKa = 10^(-5.68) - 10^(-10.8). note how in the difference taken, the higher pH represents the solution with fewer moles of H+ present per liter of solution.

So once you have moles of H+, you have moles of OH needed and then can go through NaOH molarity to find how many ml needed.

I have been trying to solve this problem since last night just like the way you described, but I'm not getting the right answer.

Moles of OH- added = Moles of H+ neutralized = 10^(-5.68) - 10^(-10.8) = 10^(-5.68)

Liters of NaOH added = 10^(-5.68) / .010 Molar = 2x10^-4 = 0.20mL

What's going on?

 

[Perpetually]

So what I omitted was that you have to take into consideration how you do not have 1 mole of Lysine. Since you have less of the AA present in solution to titrate, a lesser volume of base will need to be added to achieve the same change in pH.

 

[circulus vitios]

So what I omitted was that you have to take into consideration how you do not have 1 mole of Lysine. Since you have less of the AA present in solution to titrate, a lesser volume of base will need to be added to achieve the same change in pH.

Could you do the calculation?

 

[FishyTheFish]

Moles of H+ needed to get from first to second pKa = 10^(-5.68) - 10^(-10.8). note how in the

This is wrong. That is not the number of moles of H+ needed to get from pH of 5.68 to 10.8, that is the difference in molarity of the two solutions.


I'll post my solution, but I got a different answer:
We need to change the pH from 5.68 to 10.8. In other words, we need to reduce the [H+] concentration. We can do this by adding the NaOH. Our current molarity (moles/L) of H+ is 10^-5.68, we wish to reduce this to 10^-10.8, thus,

(10^-5.68 moles/L H+ * .01 L - .1 moles/L NaOH * x L)
--------------------------------------------------------------------------------------------------------------- = 10^-10.8
(.01L + xL)​

Where x is the volume of NaOH needed in order to get the ratio of moles H+/L soln to our desired value of 10^-10.8. We subtract .01*x in the numerator because adding x liters of NaOH will DECREASE the moles of H+.

But solving this gives an answer that is not one of the choices and I can't seem to find anything wrong with it? Anyone have any ideas of what could be wrong? On another note, the question says .01M NaOH but the answers all have .1M NaOH, which is it?

 

[greenseeking]

This is wrong. That is not the number of moles of H+ needed to get from pH of 5.68 to 10.8, that is the difference in molarity of the two solutions.


I'll post my solution, but I got a different answer:
We need to change the pH from 5.68 to 10.8. In other words, we need to reduce the [H+] concentration. We can do this by adding the NaOH. Our current molarity (moles/L) of H+ is 10^-5.68, we wish to reduce this to 10^-10.8, thus,

(10^-5.68 moles/L H+ * .01 L - .1 moles/L NaOH * x L)
--------------------------------------------------------------------------------------------------------------- = 10^-10.8
(.01L + xL)​

Where x is the volume of NaOH needed in order to get the ratio of moles H+/L soln to our desired value of 10^-10.8. We subtract .01*x in the numerator because adding x liters of NaOH will DECREASE the moles of H+.

But solving this gives an answer that is not one of the choices and I can't seem to find anything wrong with it? Anyone have any ideas of what could be wrong? On another note, the question says .01M NaOH but the answers all have .1M NaOH, which is it?

I doubled checked the book and the answers are .10M NaOH. I think you're right.. it's a typo? That would make sense if the answer was .010M.

 

[greenseeking]

Back of the book says that it takes 1 and 1/2 equivalent of NaOH to go from the 5.68 to 10.8. You can see that clearly if you draw a titration curve. But 1.5 equivalents is 15 ml so the answer should be C... but the answer key says that it's .10M of NaOH not .01M of NaOH. I think it's a typo...

 

[Perpetually]

This is wrong. That is not the number of moles of H+ needed to get from pH of 5.68 to 10.8, that is the difference in molarity of the two solutions.


I'll post my solution, but I got a different answer:
We need to change the pH from 5.68 to 10.8. In other words, we need to reduce the [H+] concentration. We can do this by adding the NaOH. Our current molarity (moles/L) of H+ is 10^-5.68, we wish to reduce this to 10^-10.8, thus,

(10^-5.68 moles/L H+ * .01 L - .1 moles/L NaOH * x L)
--------------------------------------------------------------------------------------------------------------- = 10^-10.8
(.01L + xL)​

Where x is the volume of NaOH needed in order to get the ratio of moles H+/L soln to our desired value of 10^-10.8. We subtract .01*x in the numerator because adding x liters of NaOH will DECREASE the moles of H+.

But solving this gives an answer that is not one of the choices and I can't seem to find anything wrong with it? Anyone have any ideas of what could be wrong? On another note, the question says .01M NaOH but the answers all have .1M NaOH, which is it?

Apologize for my misinformation. Glad to see it worked out.