Question

Sketch what a titration curve would look like?

1. Consider the titration of 50.00 mL of 0.01550 M NH3 with 0.03720 M HNO3. A. Sketch what a titration curve would look like for this titration, making sure to: Label both axes Illustrate a curve with the appropriate shape Indicate the location of the equivalence point on the curve (you do not need to identify its exact location on the axes) alenrennint in this titration, the titrated solution will have a pH (circle one):

Answer 1

N 4 PN - NMYND3 audit sait : NH3 250ml 0.03550 M no moles ENA = 50 LX 0.11550 1000 = 5x105 0.01550 = 7.758104 HNCE = 50 mL of 10.3720 MX10') no mole HNO3 = 50 10 3920 X10 TOOD 5x10? 0.03770 2 520.03720 x 10? = 1.867103 since the acid nuetralises equat amount of Base according to the above chemical equation . Hence Bus 7.75*10^moles arenuetralized by 775x10moles of acid Hence volume of acid actitally required for ructralising 50 me of Nin, was 50% 0.01550 = VX 0.0372 vi 50*0.0155 0.0372 VHU 3 20.83 mi. Vrul a alme. Scanned with CamScanner

we draw the titration curve let us suppose Mat it'would be a conduetometric titration since ammonia in watel furnished onione. ng + H₂O nytor the conductivity of ort is higher and thus on addition of acid it slowly decreases to the point where it is completely muetralised and further increcies аш р tw и е пf fe . and tu both scores will be different : kuause the molar conductivity of it on ions. Fonnut nost tret Nrytton" conduti Inng+Nog →quivalene pt at 20.83 ml ofte al volume of HNO3 added. Scanned with CamScanner