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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: wildareal on October 21, 2010, 09:57:53 pm

Title: Urgent Help with Stoic Titration Question
Post by: wildareal on October 21, 2010, 09:57:53 pm
A particular brand of commercially available cloudy ammonia claims not to contain more than 8% ammonia by weight.

To investigate this claim, a 21.4736g sample of cloudy ammonia was added up to a 250.0mL volumetric flask and made carefully up to the mark, with thorough mixing using distilled water. 20.00mL aliquots of this solution were titrated with 0.1030M hydrochloric acid solution. An average 22.38mL of acid was required to reach the end point, using methyl orange indicator.

Use the following questions as a guide to determining the percentage ammonia by mass in the cloudy ammonia sample:

a) Briefly explain the function of an indicator in titration.
b) Outline the difference between the endpoint and equivalence point in titration.
c) What is a standard solution?
d) Write an equation for the reaction between ammonia and hydrochloric acid: NH3 +HCL---->NH4+ +Cl-
e) Calculate the amount in mol of hydrochloric acid used per titration.
f) Calculate the amoun in mol of ammonia used per titration, and the amount in the 250.0mL volumetric falask.
g) Calculate the mass of ammonia in the 250.0mL volumetric flask.


Thanks +7
Title: Re: Urgent Help with Stoic Titration Question
Post by: luken93 on October 21, 2010, 10:20:49 pm
a) An indicator will show the point at which the titration passes the end point, to determine the average titre.
b) The equivalence point is the stage at which the molarities are equal (the both have the same number of moles), whereas the end point is the stage at which there is more of the other substance (this can usually be seen 2-3 drops after equiv point)
c) A standard solution is a solution of accurately known concentration, so it can be used accurately when titrating to find the unknown substance
d) NH3 + HCl -->  NH4+ + Cl-
e) n(HCl) = cV = (.1030) x (.02238) = 0.00230514 mol
f) n(NH3) = n(HCl) = 0.00230514 mol
g) c(NH3) = n/V = 0.00230514/.02 = 0.115257 mol/L = 0.02881425 mol/.25L = 0.02881425 in the volumetric flask

Hopefully this is correct without any errors haha