Pka/titration

[dorjiako]

What is the PH after the equivalence point has been exceeded by 5ml of the NaOH solution in the opening question.

Answer: PH= approximately 12.2
The opening question they are referring to is : Hydrazoic acid, HN3 is highly toxic compound that can cause death in minutes, if inhaled in concentrated form. 100ml of 0.2 M aqueous solution of HN3(PKA = 4.72) is to be titrated with 0.5M solution of NaOH.

Any input on how to solve this question will be appreciated!
Thanks

More so, What would be the PH of a solution that was 0.2M in HN3 and 0.10 M in N3-.

Furthermore, What would be the PH if the same amount of NaOH soulution necessary to get to the half-equivalence point in this titration were added to pure water? How does the PH of each situation compare? This demonstrates how weak acids can serve as buffer, solutions that resist changes in PH.

Any and all suggestions in tackling these questions are welcomed.

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[dorjiako]

These questions are from kaplan mcat high yield problems and I am confident that some of my SDN people are able to solve problems like this. Help still needed. All tips are welcomed.

 

[FutureDoctor503]

What is the PH after the equivalence point has been exceeded by 5ml of the NaOH solution in the opening question.

In the original problem, the volume of NaOH was determined to be 0.04 L to reach equivalence point. Add 5 mL to this volume (0.045 L). Using this volume and the concentration of NaOH, find the number of moles of NaOH (0.5*0.045=0.0225 moles). Since the number of moles of NaOH exceed the number of moles of HN3, you would have 0.0025 moles of -OH ions left in the solution (HN3 - NaOH=(0.1 L*0.2 M)-0.0225=0.0025). Now find the molarity using this value and the total volume (0.0025 mol/(0.1+0.045) L = 0.0172 M). Plug this value into the equation, pOH = - log [OH] = 1.76. To find the pH, subtract 1.76 from 14, which comes out to be 12.24.

More so, What would be the PH of a solution that was 0.2M in HN3 and 0.10 M in N3-.

This question is very simple.
For this one, you just need to use the formula, pH = pKa + log ([N3-]/[HN3]).
pH = 4.74 + log (0.10/0.20) = 4.42.

I couldn't solve the third question. 🙁

It's funny how I never did these problems before taking the actual MCAT and now, I am doing them to help others. 🙂

 

[dorjiako]

What is the PH after the equivalence point has been exceeded by 5ml of the NaOH solution in the opening question.

In the original problem, the volume of NaOH was determined to be 0.04 L to reach equivalence point. Add 5 mL to this volume (0.045 L). Using this volume and the concentration of NaOH, find the number of moles of NaOH (0.5*0.045=0.0225 moles). Since the number of moles of NaOH exceed the number of moles of HN3, you would have 0.0025 moles of -OH ions left in the solution (HN3 - NaOH=(0.1 L*0.2 M)-0.0225=0.0025). Now find the molarity using this value and the total volume (0.0025 mol/(0.1+0.045) L = 0.0172 M). Plug this value into the equation, pOH = - log [OH] = 1.76. To find the pH, subtract 1.76 from 14, which comes out to be 12.24.

More so, What would be the PH of a solution that was 0.2M in HN3 and 0.10 M in N3-.

This question is very simple.
For this one, you just need to use the formula, pH = pKa + log ([N3-]/[HN3]).
pH = 4.74 + log (0.10/0.20) = 4.42.

I couldn't solve the third question. 🙁

It's funny how I never did these problems before taking the actual MCAT and now, I am doing them to help others. 🙂

Thanks a lot for your input.

 

[indianjatt]

If the same amount of NaOH is added to the solution to get to half the equivalence point, you will need 10mmol NaOH because to reach the equivalence point you have 20mmol, so the half equivalence point is 10 mmol. This is also where pH = pKa because

HN3 + OH- --->N3- + H2O
20 mmol 10 mmol
- 10 mmol - 10 mmol
10 mmol 0 mmol 10 mmol 10 mmol
You get 10 mmol of N3- and 10 mmol of HN3.
Using henderson hassleback's equation, pH = pKa + log(a-/HA), you get pH = pKa + log(10mmol/10mmol) = pH = pKa = 4.72

This is why buffers work. They can resist a change in pH by having relatively equal concentrations of conjugate acid and base. I hope this helps.

 

[dorjiako]

If the same amount of NaOH is added to the solution to get to half the equivalence point, you will need 10mmol NaOH because to reach the equivalence point you have 20mmol, so the half equivalence point is 10 mmol. This is also where pH = pKa because

HN3 + OH- --->N3- + H2O
20 mmol 10 mmol
- 10 mmol - 10 mmol
10 mmol 0 mmol 10 mmol 10 mmol
You get 10 mmol of N3- and 10 mmol of HN3.
Using henderson hassleback's equation, pH = pKa + log(a-/HA), you get pH = pKa + log(10mmol/10mmol) = pH = pKa = 4.72

This is why buffers work. They can resist a change in pH by having relatively equal concentrations of conjugate acid and base. I hope this helps.

Thank you a lot for your great input. Would you happen to know why the answer is "The pH would be 12.9 compared to the 4.72 at the half equivalence point).
Thanks again in your explaination of this info.

 

[indianjatt]

At half equivalence point, the mmoles of base added are only half the mmoles of acid originally present, the concentrations of the conjugate base and the remaining weak acid are the same. Therefore, at half-equivalence;
pH = pKa

Idk then. I always thought that you have enough OH- to dissociate half of the weak acid and therefore you get equal concentrations of conjugate acid/base due to the reaction of 10 mmol of acid, which leads to the formation of 10 mmol of conjugate base.