Title: Chem question Post by: s.h_math on Nov 22, 2023 (https://biology-forums.com/gallery/48/1067569_22_11_23_8_14_24.png)
A student was titrating a solution of HC H O with a solution. Determine the pH at a particular point in the titration. Do this by constructing a BCA table and determining the pH. The value of Ka for HC HO is 1.5 x Complete Parts 1-2 before submitting your answer. NEXT ) 40.0 mL of a 0.200 M HC H O solution was titrated with 100.0 mL of 0.100 M (a strong base). Fill in the ICE table with the appropriate value for each involved species to determine the moles of reactant and product after the reaction of the acid and base. You can ignore the amount of water in the reaction. OH-(aq) -9 e ore 01) ange ( 01 er H20(l) + C4H702(aq) Title: Re: Chem question Post by: bio_man on Nov 22, 2023 Is this the same question with the solution?
Title: Re: Chem question Post by: s.h_math on Nov 22, 2023 unfortunately it isn't and this is due at 10pm:(.
I just have come back from the hospital and it ended up my teacher wasn't willing to offer the extra extension when in class he agreed to do so:( Title: Re: Chem question Post by: bio_man on Nov 22, 2023 I thought it was word for word the same?
Title: Re: Chem question Post by: s.h_math on Nov 22, 2023 it looks like it is but its different:(
in the after mol of HC4H7O2they have put 6.0xsomething & I don't even have that as one of the options Title: Re: Chem question Post by: bio_man on Nov 22, 2023 (https://biology-forums.com/gallery/qpics/6_22_11_23_9_32_59.png)
(https://biology-forums.com/gallery/qpics/6_22_11_23_9_33_05.png) Title: Re: Chem question Post by: s.h_math on Nov 22, 2023 (https://biology-forums.com/gallery/48/1067569_22_11_23_9_46_18.png) Its marking it as wrong unfortunately, I'm sorry but I should've provided you with the values given sir if you don't mind, could you please reply to me with what you mean by 20 mmol and 8 mmol as I'm not given those choices. in the picture below I have shown you the values I am allowed to plug in. but yes the answer is indeed correct, I have. no doubt |