Titration Lab Sheet Day 2 Alt. Need help
the blanced equation for titration 2 is NaOH + HBr ---> NaBr + H2O
So the molar ratio is 1:1
Do you know the concentration of the Acid for the second one? @wellshfella
i have to go now be back in 15 minutes
Okay no problem
Back now 2 moles of KOH react with 1 mole of H2SO4 25 mls 0.15 M KOH will reactr with 12.5 mls of 0.15 H2SO4 so concentration of H2SO4 = (12.5 / 15) * 0.150
what do you get?
gram your calculator and work it out
0.125?
Wait. That's for Titration #1, I meant for # 2
Can you help me with that one?
oh right
Hold on the volumes in the table are different form the ones in the question. Which are correct?
Oh right, well you see they are the same measurements but jus with a different unit. Instead of being ML they are L. But the teacher said to change the ml to L so thats what I did.
They are the same amount. Use the ones on the table.
yes but they are still different. I guess the ones in question are correct?
Well yea they are different. Whichever one you prefer I guess. As long as we get the same answer out of both
in Q NaOH = 0.02l in T Na OH = 0.03L
Ok so we'll take the ones in the question 0.02L of 0.250 M NaOH will neutralise 0.02L of 0.250 M HBr ( because molar ratio = 1:1) but the volume of acid = 0.03L which means that the HBr is weaker so concentration of the acid is 2/3 * 0.250
0.17 (rounded to the nearest hundredth)
yes
So the concentration of the acid would be 0.17 M?
to be honest I'm confused by the question and table being different but thats the answer if our assumptions are correct.
Well the teacher helped us out in the first one but since I had to leave class early I didnt get to see the whole thing. And she used the L format to find the concentration.
OH - its the concentration of the base they want in the question!!
im getting a headache! lol but i have to go..
Lol i am too. Im as lost as i can get. Idk, i think ill jus leave it as blank.
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