Please answer the following questions regarding the titration curve shown below.

Question

Please answer the following questions regarding the titration curve shown below. 14 12 10 PH 0 10 15 20 25 30 35 40 45 50 55 60 mL titrant added (a) Choose which of the following is true about the analyte in this titration O It is a monoprotic acid. O It is a diprotic acid. O It is a triprotic acid O It is a monoprotic base. O It is a diprotic base. O It is a triprotic base. O None of the above. (b) If the titrant has a molarity of 0.1750 M and there are 45.00 mL of analyte present, what is the molarity of the analyte? (c) If 0.4590 g of analyte were dissolved in 45.00 mL of water to perform this titration, what is the molecular mass of the analyte? 187.43 ] X g/mol

Explanation / Answer

B) B) This question involves titration curve,

now look for the equivalence point /steep st line ..in the curve where moles of titrant is = to moles of base .
Here its at - 20 ml (base/titrant) then M of the anylate can we deduced like this
Then apply M1V1 = M2V2.
-Ma = (20) (.1750) /45
=.0778.

C) we know that at 45 mL molarity of analyte is 0.778.

So mol = conc x volume

And mol = given weight/ molecular mass.

Mol = 0.0778 M x (.02 + .045) L = 0.00506

Molecular mass = 0.459 g / 0.00506 mol = 90.8 g/mol

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