Thank you so much! This broke down a lot of the simple things that my professor kept using without explanation (like why we couldn't use M in an ice table in a titration, for example), which has been frustrating me throughout this unit
Glad it helped! These weak acid - strong base titration problems are pretty tricky and there are soooo many details to keep straight. If you're working on this topic, you're almost done with all the nasty ICE table stuff - it's not going to get any worse than what you've already done. Keep grinding, you're almost through it.
This might be a dense question, but why can't we use (CH3)3NH+ to calculate pH, given that it's the conjugate acid and source of H+ ions on the right side of the equation? Is it normal that you use whatever strong acid is left to calculate pH? I guess I thought that you could only use what's on the right side of the equation after everything's reacted, if that makes sense.
Excellent question - this is confusing stuff. Yes, if you have any strong acid (or strong base), it will completely dictate the pH of the solution and you should ignore any contributions from a weak acid or base. Because strong acids (and bases) completely dissociate, they will always have a more significant contribution to the pH of the solution and essentially overpower any contribution that might come from a weak acid or base.
@@RoxiHulet Okay, I'll just make sure to sear that into my brain haha. Thank you so much! I really appreciate you taking the time to answer me! You're saving so many people's sanity with your videos
This ICE is done in moles, as stated in the video. You cannot use molarity in an ICE table used for a titration calculation due to the volume change associated with titrations.
0.005774 is the number of moles of acid, not the molarity. pH is calculated using the molarity of the strong acid, so your method would give the wrong answer. This is actually explained in detail the video, so I am puzzled by your comment.
Thank you so much! This broke down a lot of the simple things that my professor kept using without explanation (like why we couldn't use M in an ice table in a titration, for example), which has been frustrating me throughout this unit
Glad it helped! These weak acid - strong base titration problems are pretty tricky and there are soooo many details to keep straight. If you're working on this topic, you're almost done with all the nasty ICE table stuff - it's not going to get any worse than what you've already done. Keep grinding, you're almost through it.
You're the biggest life saver Roxi!
This might be a dense question, but why can't we use (CH3)3NH+ to calculate pH, given that it's the conjugate acid and source of H+ ions on the right side of the equation? Is it normal that you use whatever strong acid is left to calculate pH? I guess I thought that you could only use what's on the right side of the equation after everything's reacted, if that makes sense.
Excellent question - this is confusing stuff. Yes, if you have any strong acid (or strong base), it will completely dictate the pH of the solution and you should ignore any contributions from a weak acid or base. Because strong acids (and bases) completely dissociate, they will always have a more significant contribution to the pH of the solution and essentially overpower any contribution that might come from a weak acid or base.
@@RoxiHulet Okay, I'll just make sure to sear that into my brain haha. Thank you so much! I really appreciate you taking the time to answer me! You're saving so many people's sanity with your videos
my HH equation did not work for some reason, and your work is very different from Aleks explanation.
It is not unusual for my work to be very different from the ALEKS explanation. There is often more than one way to approach a problem :)
did you get that coz i followed your method and got a wrong
you do not divide the strong acid by the total volume since an ICE table is molarity not moles.
This ICE is done in moles, as stated in the video. You cannot use molarity in an ICE table used for a titration calculation due to the volume change associated with titrations.
The correct answer is given by using -log(0.005774)
0.005774 is the number of moles of acid, not the molarity. pH is calculated using the molarity of the strong acid, so your method would give the wrong answer.
This is actually explained in detail the video, so I am puzzled by your comment.