Question

The pH of 0.5 L of 1.0 M NaCl after the electrolysis for 965 s using 5.0 A current, is .
.....?

Answer 1

Electrolysis of aqueous NaCl gives following products:

At cathode:

Now,

Moles of NaCl = molarity x volume

= 1 x 0.5

= 0.5

Amount of charge = current x time

= 5 x 965 C

Moles of electrons passed = Q/F

= (5 x 965) / 96485

= 0.05

According to the equation,

2 moles of NaCl produce 2 moles of OH- ions.

0.5 moles of NaCl will produce 0.5 moles of OH- ions.

But the number of moles of electrons passed = 0.05

So, number of moles of OH- formed = 0.05

Concentration of OH- ions in 0.5 L = 0.05/0.5 = 0.1 M

Now,

pOH = -log[OH-] = -log[0.1] = 1

pH = 14 00 - pOH = 14 – 1 = 13.

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