2.50g of an unknown carbonate were dissolved in 100cm3 of 1.00moldm−3 hydrochloric acid (an excess). The resulting solution was made up to 250cm3 in a volumetric flask. 25.00cm3 aliquots of this solution were titrated against 0.250moldm−3 sodium hydroxide. The titre is 24.10cm3. I've calculated amount of NaOH as 0.006025 mol. Calculate the amount of hydrochloric acid remaining after reaction.
This was a 25cm3 sample from 250cm3, so you must have 10 times as many moles HCl in the entire 250cm3 flask as were in the 25cm3 sample pipetted and titrated.
This was a 25cm3 sample from 250cm3, so you must have 10 times as many moles HCl in the entire 250cm3 flask as were in the 25cm3 sample pipetted and titrated.
I calculated the moles in in a sample as 0.01, so I then subtracted 0.006025 from that. And it's not right?
That was the concentration before the unknown carbonate was added. The carbonate reacted with some of the HCl.
The titration with NaOH is to find how much remains unreacted after the carbonate. The titration with NaOH will tell you the number of moles of HCl in the 25cm3.
You just need to have a balanced equation of NaOH and HCl. The titration must obey that equation, and you know the number of moles of NaOH as you already stated that.
How many moles of HCl in the 25cm3 sample? So how many in the 250cm3 flask?
That was the concentration before the unknown carbonate was added. The carbonate reacted with some of the HCl.
The titration with NaOH is to find how much remains unreacted after the carbonate. The titration with NaOH will tell you the number of moles of HCl in the 25cm3.
You just need to have a balanced equation of NaOH and HCl. The titration must obey that equation, and you know the number of moles of NaOH as you already stated that.
How many moles of HCl in the 25cm3 sample? So how many in the 250cm3 flask?
Ah, I understand now. The question is asking for the amount of HCl after the carbonate is added, which would be 10x0.00625. Thanks for the help!