Question

a 405- mL soution of NaCl was electrolyzed for 8.00 min. If the pH of the final solution was 12.09 calculate the average current used. _____A

Answer 1

aquoes NaCl gives OH- at cathode

we get [OH-] by using pH , pH = 12.09 , pOH = 14-pH = 14-12.09 = 1.91

[OH-] = 10^ -pOH = 10^ -1.91 = 0.0123 M

OH- moles = M x V ( in L) = 0.0123 x ( 405/1000) = 0.0049826

number of electrons involved = number of moles of OH- x charge of OH-   = 0.0049826 x 1 = 0.0049826 moles

Number of columbs of charge = Moles x Farady constant = 0.0049826 x 96485 = 480.746 C

we have formula charge q = current x time in secn

      480.746 = i x ( 8 x 60)

i = 1 A