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NotTim
 one year ago
ADSORPTION OF ACETIC ACID BY ACTIVATED CHARCOAL LAB"Calculate the exact concentration of your base. To do this, use the volume of base added to 10 mL of your standardized acid solution. Calculate the base concentration for each titration performed then determine the mean concentration."
NotTim
 one year ago
ADSORPTION OF ACETIC ACID BY ACTIVATED CHARCOAL LAB"Calculate the exact concentration of your base. To do this, use the volume of base added to 10 mL of your standardized acid solution. Calculate the base concentration for each titration performed then determine the mean concentration."

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NotTim
 one year ago
Best ResponseYou've already chosen the best response.0I have titration volumes of 11.72, 9.49, 10.27, 10.47, 10.27 mL.

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0MY Potassium Hydrogen Phthalate concentration is 0.1024 M

NotTim
 one year ago
Best ResponseYou've already chosen the best response.01. Freundlich Isotherm 2. Langmuir Isotherm are mentioned in the lab intro.

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0ITs like this lab: http://www.cpp.edu/~sjanz/physical_chemistry/chm_353_labs/adsorption_of_acetic_acid_by_a_solid.pdf

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0http://www.isca.in/rjcs/Archives/vol2/i9/9.ISCARJCS2012145.pdf The proper research article, surprisngly not helpful though.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is the acid ur titrant or ur base?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for ur first one, 11.72 ml of ur base was used to neutralise the acid?

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0Wait base is the thing that is the beaker already right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ive forgotten too haahah, its been a yr since i did titration in uni

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but i think you can have either acid or base as ur titrant, it just depends what ur experiment tells u, i think the norm is to have the base in the beaker thought (from my vauge memory)

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0Alright, any ideas on figuring out concentrations?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0best bet is dilution factor c1v1=c2v2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0give me a sec, i'm working on a maths problem as we speak, i posted the question and need some help haha

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0i dunno what to use for v1 and v2 though i just know that i titrated 11.72 mL into 10mL KHP.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0KHP is ur acid right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so you titrated 11.72ml of base into 10ml KHP?

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0fr my first sample, yes.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait, dilution factor is unessecary, do you know the reaction equation?

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0no, i don't remember what that means

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i reckon that the key

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0as in do you know the stoichiometry is it 1:1?

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0I'm looking thru the lab manual, and it says nithing.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cause thats sorta what u wanna do, find moles etc then use stoichiometry to find ur base conc

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0you think i was suppose to deriive that from the equation?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what is ur base ur using?

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0sorry, was at bathrooom. its naoh

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well that makes life so much easier ahah

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0CH3COOH + NaOH > CH3COONa + H2O

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so stoichiometry is 1:1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0C(Acetic)= 0.1024 M V(acetic)=10ml V(naoh)=11.72 C(naoh)=??

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0acetic is the part after this section of the lab

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[KHC _{8}H _{4}O _{4}+NaOH \rightarrow H _{2}O+NaKC _{8}H _{4}O _{4}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0same idea C(KHP)= 0.1024 M VKHP)=10ml V(naoh)=11.72 C(naoh)=??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0C(KHP)= 0.1024 M VKHP)=0.01L V(naoh)=0.01172L C(naoh)=??M

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0lol im stupid i was trying to figure out how you got 0.01172 L...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0n(kHp)=0.1024*0.01=0.001024 moles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we know mole ratio is 1:1 therefore number of acid moles=number of base moles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0n(naoh)=0.001024 moles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0n=CV C=n/V C(NaOH)=0.001024/0.01172 C(NaOH)=0.087M

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0i don't know where n=cv is coming from; as in, why its being applied to this scenario

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because u want to find the base concentration...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and you can only do that using stoichiometry of what u you (acid)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 11.72, 9.49, 10.27, 10.47, 10.27 mL. Test 1: C(naoh)=0.087M Test 2: C(naoh)= 0.001024/(9.49/1000)=0.108M Test 3: C(naoh)= 0.001024/(10.27/1000)=0.0997M Test 4: C(naoh)= 0.001024/(10.47/1000)=0.0978M test 5: C(naoh)= 0.001024/(10.27/1000)=0.0997M Therefore mean concentration is the average right. C(naoh)= test 1+test2+test3+test4+test5/5runs C(naoh)=0.098M (to be honest, no lab supervisor would make up batches of 0.098M, we can safely say that the sodium hydroxide concentration is 0.1M

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0You just had to explain up to n=CV. But thanks!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its because moles of a reaction always is constant, its just concentration changes. so we use stoichiometry (which is in effect molar ratio) so we use a comparison of moles to determine unknown concentrations
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