Question
A CURRENT OF 0.25 A IS PASSED THROUGH 400ML OF A2.0 M SOLN OF NaCl FOR 35 MINUTES.WHAT WILL BE THE PH OF THE SOLN AFTER THE CURRENT IS SWITCHED OFF?
Joshi sir comment
q = it = 0.25*35*60 = 525 coul.
so no. of eq.( of H2) formed = 525/96500 = 21/3860, same will be the no. of moles
total initial no. of moles of NaCl = 2*0.4 = 0.8 moles
remaining no. of moles of H+ = 0.8 - (21/3860)
so pH = -log[H+]