chem: PH.. not even possible?

[ADentalHopeful]

ok.
so you have 2.8 x 10^-4 M HCl
and you have 2.84x10^-4 M NaCl in a solution.
what is the ph of this solution?

HCl is a strong acid so no ka value.
but, how do you find the PH of this solution without an experiment? and is this even possible..

so confused.
I always assumed no ka no calculation.
cause i want to make an ice table right. and then get like
(2.84x10^-4 +x)(x)/(2.8x10^-4)=ka
seems quite not do-able to me...

oh and if you add .001 mol NaOH what is the PH...
like these should be easy.
I think it is a trick

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[Nimsi Nims]

ok.
so you have 2.8 x 10^-4 M HCl
and you have 2.84x10^-4 M NaCl in a solution.
what is the ph of this solution?

HCl is a strong acid so no ka value.
but, how do you find the PH of this solution without an experiment? and is this even possible..

so confused.
I always assumed no ka no calculation.
cause i want to make an ice table right. and then get like
(2.84x10^-4 +x)(x)/(2.8x10^-4)=ka
seems quite not do-able to me...

oh and if you add .001 mol NaOH what is the PH...
like these should be easy.
I think it is a trick

I think you don't even need an ICE table. just take the negative log of the concentraion of HCl because that's the only source of H+ in you solution. So I'd say the pH would be 3.6. Let me know If I have the right answer.

 

[Nimsi Nims]

For the second part I would suggest using the table for neutrilization.

 

[Nimsi Nims]

I got the pH 10.9 after adding of .001 mol NaOH. Let me know if this is right !!!

 

[joonkimdds]

I think you don't even need an ICE table. just take the negative log of the concentraion of HCl because that's the only source of H+ in you solution. So I'd say the pH would be 3.6. Let me know If I have the right answer.

that's what i think and what i got too~

 

[PooyaH]

ok.
so you have 2.8 x 10^-4 M HCl
and you have 2.84x10^-4 M NaCl in a solution.
what is the ph of this solution?

HCl is a strong acid so no ka value.
but, how do you find the PH of this solution without an experiment? and is this even possible..

so confused.
I always assumed no ka no calculation.
cause i want to make an ice table right. and then get like
(2.84x10^-4 +x)(x)/(2.8x10^-4)=ka
seems quite not do-able to me...

oh and if you add .001 mol NaOH what is the PH...
like these should be easy.
I think it is a trick

NaCl is a salt and it would dissociate in the solution. Na+ cation is associated with a strong base (NaOH) and Cl- anion is associated with a strong acid (HCl) so it would have no effect on the pH! so since we have 2.8x10^-4 M of HCl the pH should be slightly lower than 4 (between 3 & 4)! NaCl is there to mislead you!

If you add 1x10^-3 M of NaOH would be slightly basic I think since we .001 M is higher than 2.8x10^-4 M of HCl!

 

[ADentalHopeful]

thanks all of you for your help!
that was my hunch, but i wasnt sure if you could do that....