In an experiment 0.04F was passed through 400ml of 1M NaCl solution . What would be PH of the solution after electrolysis

A_8. B_10. C_13. D_6

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Solution

(C_13)

At electrolysis we get: NaCl + H2O --> NaOH (at anode) + 1/2H2 + 1/2Cl2 (at cathode)
Now, for 1M solution of NaCl, the no. of moles = molarity x volume in L = 1 x 0.4L = 0.4 mole
Since, 1F produces = 1 mole of NaOH
So, 0.04F will produce = 0.04 mole of NaOH
So, 400 ml of the solution contains = 0.04 mol NaOH
Hence, molar concentration of NaOH = 0.04/0.4L = 0.1 M
Hence. [OH-]=0.1M
pOH = -log[OH-] = 1
Hence.,
pH = 14-1 = 13

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In an experiment 0.04F was passed through 400ml of 1M NaCl solution . What would be PH of the solution after electrolysis A_8. B_10. C_13. D_6

In an experiment 0.04F was passed through 400ml of 1M NaCl solution . What would be PH of the solution after electrolysis

A_8. B_10. C_13. D_6