[Technicalhuman]

Usually when we have a pH calculation question, the first part would ask us to find the pH of pure NH3. So we would use the ICE table to find new concentration of NH4+ and OH-. However, in the next part they would say some acid like HCl was added so we would use NH3+HCl with the initial number of moles of NH3 to be the given concentration times volume instead of the new concentration of NH3 found in the first part.

Why is this so? Shouldn't the concentrations used be the new ones?

Thanks

[Borek]

Please elaborate, I have problems understanding what you mean. Give an example of the complete question and calculations.

[Technicalhuman]

Please elaborate, I have problems understanding what you mean. Give an example of the complete question and calculations.

Hi Borek thanks for the reply.

The question is 25ml of 0.25M NH3 is titrated with 0.25M HCl. What is the (a) Initial pH (b) Final pH when 10ml of HCl was titrated.

So for (a) I would just draw out an ICE table for the reaction NH3+H2O NH4+ +OH- and get the pOH using the Kb value of NH3 hydrolysis. After that I'd take 14-pOH to get the pH.

But for part (b) in guide books and notes, they simply take 0.25 x25/1000 to get the total number of moles of NH3 but I don't quite understand this because the concentration of NH3 changes from part (a). Asking my sister, she said that we should use the 0.25M concentration because its actually the OH- from the hydrolysis of the NH3 which is reacting with the HCl so as we add in the HCl, the hydrolysis of NH3 is driven forward producing more NH4+ and OH-. However she said this was just a guess of hers. So i was wondering what is the actual reason for this.

Thanks for the help.

[Borek]

they simply take 0.25 x25/1000 to get the total number of moles of NH3

This IS the total number of moles of ammonia. But it is not yet the final answer.

[Technicalhuman]

they simply take 0.25 x25/1000 to get the total number of moles of NH3

This IS the total number of moles of ammonia. But it is not yet the final answer.

Hi Borek thanks for the reply.

I don't quite get why that is the total number of moles as from (a) its shown that the concentration of ammonia decreases a little. But when i continue with the question, NH3+HCl --> NH4Br and I would minus 0.25x25/1000 with 0.25x10/1000 to get the number of moles of NH3. While for NH4+, the number of moles would just be 0.25x10/1000. However, I'm still not sure why the 0.25x25/1000 is the actual number of moles of ammonia since there is already some equilibrium that causes some NH3 to be hydrolyzed.

Thanks Borek.

[Borek]

Looks like you are mixing total number of moles of ammonia and ammonium ions with equilibrium concentrations of both.

To calculate pH at the end of the titration you should calculate theoretical analytical concentration of NH₄⁺ (ie assume all ammonia was protonated and calculate its concentration remembering original solution was diluted with the titrant), then treat the solution as solution of a weak acid NH₄⁺ and calculate pH using ICE table.

[Technicalhuman]

Looks like you are mixing total number of moles of ammonia and ammonium ions with equilibrium concentrations of both.

To calculate pH at the end of the titration you should calculate theoretical analytical concentration of NH₄⁺ (ie assume all ammonia was protonated and calculate its concentration remembering original solution was diluted with the titrant), then treat the solution as solution of a weak acid NH₄⁺ and calculate pH using ICE table.

Ho borek thanks again for the reply.

What do you mean by treating all the NH3 to be protonated? Isn't it the NH3 reacting with the HCl? And even so why can we just assume all the NH3 to be protonated? Since in part A we already found out that only a small amount of NH3 actually gets protonated.

Thanks again for the help.

[magician4]

you could work out the "very exact" pH values for those problems by using more or less complex approaches, like a charge balance, for example

the results gained thereof , however, will be very close (i.e. for all practical purposes, like calculating the resulting pH thereof with "reliable" 2 decimals ) identical to approximations - but much more difficult to calculate

what you would do here instead would be,  to calculate with approximation equations like "pOH = 0.5 (pK_b - log c₀ ) " (for the initial pH of a weak base), or "consumption of deficit HCl is considered complete, with 1:1 forming of NH₄⁺ thereof" (for reaction of HCl with NH₃ ) or "NH₄⁺ formed from HCl will be all the NH₄⁺ there is" (neglection of further self-dissociation of NH₃ , compared to this) , respectively.

this way, you can work out the final pH from the Henderson-Hasselbalch eq. easily

regards

Ingo

[Technicalhuman]

you could work out the "very exact" pH values for those problems by using more or less complex approaches, like a charge balance, for example

the results gained thereof , however, will be very close (i.e. for all practical purposes, like calculating the resulting pH thereof with "reliable" 2 decimals ) identical to approximations - but much more difficult to calculate

what you would do here instead would be,  to calculate with approximation equations like "pOH = 0.5 (pK_b - log c₀ ) " (for the initial pH of a weak base), or "consumption of deficit HCl is considered complete, with 1:1 forming of NH₄⁺ thereof" (for reaction of HCl with NH₃ ) or "NH₄⁺ formed from HCl will be all the NH₄⁺ there is" (neglection of further self-dissociation of NH₃ , compared to this) , respectively.

this way, you can work out the final pH from the Henderson-Hasselbalch eq. easily

regards

Ingo

Hi thanks for the reply.

Is the NH3 the reactants that is reacting? Or is it the NH4+ and OH- that is reacting with the HCl? Because my sister explained that it's not right to say that the NH3 is reacting.

If so, then can't we use the concentration found in part A to solve rather than to use the given concentration?

thanks for the help