0.25 A current is passed for 35 min through 400ml 2.0 M NaCL solution, what will be pH of the solution?
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Answer:
Nacl→Na
+
+cl
−
H
2
O→H
+
+OH
−
at anodes cl
−
will get oxidized
2cl
−
→cl
2
+2e
−
At cathode, H
+
will be reduced
2H
+
→
2e
−
H
2
(g)
Q=I×t=(9.65)(100)=965 coulombs
965C=0.01F
(∵IF=96500 coulombs)
moles of Nacl =
n
M
Eq.wt
mass
=
96500
Q
=F
0.01F=
Eq.wt
mass
[n
moles
=0.01]
morality
100
0.01
×1000=0.1M
[OH
−
]=0.1H
POH=1
PH=14−1=13.
∴ PH of resulting solution =13.
Explanation:
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