During a titration, the following data were collected. A 25mL portion of an unknown solution was titrated with 1.0 M NaOH. It required 65mL of the base to neutralize the sample. How many moles of acid are present in 3.0 liters of this unknown solution?
2006-09-22 06:51:04 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
M(acid)V(acid) = M(base)V(base), so the molarity of your acid is (65)*1/25 = 2.6M or 2.6 moles/L.
3 liters would have 7.8 moles of acid. (Not sure where the hell people got 56, unless they're just being assholes.)
2006-09-22 07:00:13 · answer #1 · answered by TheOnlyBeldin 7 · 0⤊ 0⤋
56
2006-09-23 01:42:22 · answer #2 · answered by Anonymous · 0⤊ 0⤋
56
2006-09-22 06:52:00 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Your unkown look a strong acid,
25xN1 = 65x1 N1,is the normality of the acid
if your acid is a monobasic acid it will be its molarity also
M1=65/25 =2.6 molar
3x2.6 =7.8 moles
2006-09-22 07:04:50 · answer #4 · answered by basimsaleh 4 · 0⤊ 0⤋
56?
2006-09-22 06:52:59 · answer #5 · answered by gooterscooby 3 · 0⤊ 0⤋
(M)(Liters)=moles
1.0M x 0.065L = 0.065moles
M = mol/L so: 0.065mol/0.025L = 2.6mol/L or 2.6M
This holds true only if the mol ratio of acid and base is a one to one mol ratio.
2006-09-22 17:53:30 · answer #6 · answered by r.stolk 2 · 0⤊ 0⤋
10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 trillions ...hahhahahahah
THAT WAS HARD
2006-09-22 06:53:06 · answer #7 · answered by kenneth x 1 · 0⤊ 0⤋
We're not here to do your homework for you
2006-09-22 06:52:27 · answer #8 · answered by Anonymous · 0⤊ 0⤋