JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ec214
50.0mL of 0.0200M NaCl is titrated conductometrically with 0.200MAgNOX3. Limiting molar ionic conductivities (in Scm2mol−1): NaX+=50, ClX−=76, AgX+=62, NOX3X−=71. Ignore the volume change and the concentration dependence of ionic conductivities. The correct statement(s) is/are
AThe initial conductivity of the NaCl solution is 2.52mScm−1
BUp to the end point the conductivity decreases slightly, because ClX− is replaced by the less mobile NOX3X−
CThe end point is reached at 5.00mL of AgNOX3
DAfter the end point the conductivity increases per mL of titrant more slowly than it decreased per mL before the end point
Answer: A, B, C
Initially κ=(50+76)×0.0200/1000=2.52×10−3Scm−1, so (A) is correct. Before the end point AgX++ClX−AgCl(s) removes ClX− (λ∘=76) and leaves NOX3X− (λ∘=71) in its place, a small decrease of 5×0.0200/1000=0.10mScm−1 overall, so (B) is correct. ClX− present =1.00mmol, needing 1.00/0.200=5.00mL of AgNOX3, so (C) is correct. Each mL of titrant adds 0.200 mmol, i.e. 0.00400 M in 50 mL: before the end point Δκ=−(76−71)(0.00400)/1000=−0.020mScm−1 per mL, after it Δκ=+(62+71)(0.00400)/1000=+0.532mScm−1 per mL, so the rise is much faster and (D) is wrong.