Calculate pCl for the titration of 200 ml 0.2 M NaCl with 0.2 MAGNO, for the addition of 0.0, 20.0, 99.0, 99.5, 200.0 and 210.0 ml AgNO3
K, for AgCl is 1.0 x10-10
Answers
Answer:
Explanation:ANSWER
No. of moles of Ag
+
ions =200×
1000
0.005
=10
−3
moles
No. of moles of Cl
−
ions =300×
100
0.01
=3×10
−3
moles
Ag
(aq)
+
+Cl
(aq)
⊖
→AgCl(s)
Cl
−
ions are in excess, therefore, it will consume all Ag
+
ions and form AgCl
Thus, Amount of Cl
−
ions left =3×10
−3
−10
−3
=2×10
−3
moles
[Cl
−
]
left
=
300+200
2×10
−3
×1000=4×10
−3
M=0.004 M
Dissociation of AgCl to its ions will produce an equal amount of each ion ( say S)
Then [Ag
+
]=S and [a
−
]=S+0.004≃0.004
K
cp
=[Ag
+
][Cl
−
]=1.8×10
−10
⇒[Ag
+
]=
[Cl
−
]
1.8×10
−10
=
0.004
1.8×10
−10
Maximum conc of Ag
+
in the mixture is 4.5×10
−8
M
Hence, the correct option is B